sLeetCode笔记

10-10 LeetCode_9回文数

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 // 方法1：将整数转为字符串,前后指针循环public boolean isPalindrome(int x) {//1.判断x<0 负数不肯是回文数if (x < 0) {return false;}//2. 转化字符串String s = x + "";//3.定义左右指针int left = 0;int right = s.length() - 1;while (left < right) {if (s.charAt(left++) != s.charAt(right--)) {return false;}}return true;}

// 数字最大数21亿反转71亿public boolean isPalindrome3(int x) {// 特殊情况：// 如上所述，当 x < 0 时，x 不是回文数。// 同样地，如果数字的最后一位是 0，为了使该数字为回文，// 则其第一位数字也应该是 0// 只有 0 满足这一属性// x%10 取一个，不要一个x/10if (x < 0 || (x % 10 == 0 && x != 0)) {return false;}int revertedNumber = 0;while (x > revertedNumber) {revertedNumber = revertedNumber * 10 + x % 10;x /= 10;}// 当数字长度为奇数时，我们可以通过 revertedNumber/10 去除处于中位的数字。// 例如，当输入为 12321 时，在 while 循环的末尾我们可以得到 x = 12，revertedNumber = 123，// 由于处于中位的数字不影响回文（它总是与自己相等），所以我们可以简单地将其去除。return x == revertedNumber || x == revertedNumber / 10;}

10-11 LC20 有效地括号

class Solution {public boolean isValid(String s) {if (s == null || s.length() == 0) {return false;}//1.把字符串转成字符数组char[] str = s.toCharArray();//2.声明一个StackStack<Character> stack = new Stack<>();//3.遍历配对for (int i = 0; i < str.length; i++) {char cha = str[i];if (cha == '(' || cha == '[' || cha == '{') {//3.1  如果是3个括号，配对stack.add(cha == '(' ? ')' : (cha == '[' ? ']' : '}'));} else {//3.2 如果stack为空，不对if (stack.isEmpty()) {return false;}//3.3 不为空，说明配对成功，弹出char last = stack.pop();if (cha != last) {return false;}}}return stack.isEmpty();}}

10-11 LC20 有效地括号（法2）

https://www.bilibili.com/video/BV1J3411n7kU?p=2

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10-13 LC13罗马To数字

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class Solution {public  int romanToInt(String s) {//1. 判断长度为0if (s.length() == 0) {return 0;}//2声明Map并赋值Map<Character, Integer> map = init();//3取出第一个字符int  sum = map.get(s.charAt(0));//4.比较大小for (int i = 1; i <s.length() ; i++) {char curRoman=s.charAt(i);int curInt=map.get(curRoman);char lastRoman=s.charAt(i-1);int lastInt=map.get(lastRoman);sum+=curInt;if(curInt>lastInt){sum=sum-2*lastInt;}}return sum;}public  Map<Character, Integer> init() {Map<Character, Integer> map = new HashMap<>();map.put('I', 1);map.put('V', 5);map.put('X', 10);map.put('L', 50);map.put('C', 100);map.put('D', 500);map.put('M', 1000);return map;}
}

10-14 冒泡排序

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package day04;import java.util.Arrays;//冒泡排序
public class BubbleDemo {public static void main(String[] args) {//int[] arr = {8,4,2,1,23,344,12};int[] arr = {4,6,3,7,8,9,10};//未优化//sort1(arr);//优化1 针对比较轮数进行优化//sort2(arr);//优化2 针对比较次数进行优化sort3(arr);}//未优化 冒泡排序public static void sort1(int[] arr) {for (int i = 0; i <arr.length-1; i++) {   //i 轮数System.out.println("目前是第"+(i+1)+"轮比较");for (int j = 0; j <arr.length-1-i ; j++) {   //j 次数System.out.println("第"+(i+1)+"轮，第"+(j+1)+"次比较");//交换的条件if(arr[j] > arr[j+1]){int temp = arr[j+1];arr[j+1] = arr[j];arr[j] = temp;}}//输出当前轮次比较的结果System.out.println(Arrays.toString(arr));}//输出System.out.println("排序后："+Arrays.toString(arr));}//优化1 针对比较轮数进行优化 冒泡排序public static void sort2(int[] arr) {for (int i = 0; i <arr.length-1; i++) {   //i 轮数System.out.println("目前是第"+(i+1)+"轮比较");//标志位boolean flag = true;for (int j = 0; j <arr.length-1-i ; j++) {   //j 次数System.out.println("第"+(i+1)+"轮，第"+(j+1)+"次比较");//交换的条件if(arr[j] > arr[j+1]){int temp = arr[j+1];arr[j+1] = arr[j];arr[j] = temp;flag = false;}}//输出当前轮次比较的结果System.out.println(Arrays.toString(arr));//当不再发生交换时，则结束比较if(flag){break;}}//输出System.out.println("排序后："+Arrays.toString(arr));}//优化2 针对比较次数进行优化 冒泡排序public static void sort3(int[] arr) {//定义一个作为比较的边界int position = arr.length-1;for (int i = 0; i <arr.length-1; i++) {   //i 轮数System.out.println("目前是第"+(i+1)+"轮比较");//标志位boolean flag = true;//记录最后交换位置的值int limit = 0;for (int j = 0; j <position ; j++) {   //j 次数System.out.println("第"+(i+1)+"轮，第"+(j+1)+"次比较");//交换的条件if(arr[j] > arr[j+1]){int temp = arr[j+1];arr[j+1] = arr[j];arr[j] = temp;flag = false;limit = j;}}position = limit;//输出当前轮次比较的结果System.out.println(Arrays.toString(arr));//当不再发生交换时，则结束比较if(flag){break;}}//输出System.out.println("排序后："+Arrays.toString(arr));}
}

10-14 二分查找

package day04;import java.util.Arrays;
import java.util.Scanner;//练习2：有一个数列：8,4,2,1,23,344,12，此时从键盘中任意输入一个数据，判断数列中是否包含此数
//方法二：先冒泡排序，再折半查找出是否包含
public class BinaryDemo {public static void main(String[] args) {int[] arr = {8,4,23,2,1,23,344,12};//冒泡排序for (int i = 0; i <arr.length-1; i++) {   //i 轮数for (int j = 0; j <arr.length-1-i ; j++) {   //j 次数//交换的条件if(arr[j] > arr[j+1]){int temp = arr[j+1];arr[j+1] = arr[j];arr[j] = temp;}}}//输出System.out.println("排序后："+Arrays.toString(arr));//二分搜索法Scanner scanner = new Scanner(System.in);System.out.println("请输入搜索数据：");int key = scanner.nextInt();int left = 0;int right = arr.length-1;int middle = 0;boolean flag = true;while (left <= right){middle = (left + right)/2;if(key > arr[middle]){//看右边left = middle + 1;}else if(key < arr[middle]){//看左边right = middle - 1;}else if(key == arr[middle]){flag = false;System.out.println("存在");break;}}if(flag){System.out.println("不存在");}}}

10-14 LC12数字转罗马

    public static String intToRoman(int num) {//I             1//V             5//X             10//L             50//C             100//D             500//M             1000String[] qian = {"", "M", "MM", "MMM"};String[] bai  = {"","C","CC","CCC","CD","D","DC","DCC","DCCC","CM"};String[] shi  = {"","X","XX","XXX","XL","L","LX","LXX","LXXX","XC"};String[] ge={"","I","II","III","IV","V","VI","VII","VIII","IX"};return qian[num/1000] +bai[num/100%10]+shi[num%100/10]+ge[num%10];}

优化版

class Solution {int[] values = {1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1};String[] symbols = {"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"};public String intToRoman(int num) {StringBuffer roman = new StringBuffer();for (int i = 0; i < values.length; ++i) {int value = values[i];String symbol = symbols[i];while (num >= value) {num -= value;roman.append(symbol);}if (num == 0) {break;}}return roman.toString();}
}

LC 704. 二分查找

package day0;public class LC704_binarySerach {public int search(int[] nums, int target) {//给定一个 n 个元素有序的（升序）整型数组 nums 和一个目标值 target  ，写一个函数搜索 nums 中的 target，如果目标值存在返回下标，否则返回 -1int left = 0;int right = nums.length - 1;while (left <= right) {int mid = (left + right) / 2;if (target == nums[mid]) {return mid;} else if (target > nums[mid]) {left = mid + 1;} else {right = mid - 1;}}return -1;}}

LC 278. 第一个错误的版本

package day0;public class LC278_binarySeracher第一个错误的版本 {public int firstBadVersion(int n) {int left = 1, right = n;while (left < right) { // 循环直至区间左右端点相同int mid = left + (right - left) / 2; // 防止计算时溢出if (!isBadVersion(mid)) {// 不是坏的，就是好的// 1好的  2好的 3好的 4坏的  5坏的return left=mid+1;} else {// 就是坏的 继续向前面看，有没有好的,，不能-1 如果3是坏的 2也是坏的 （1+2）/2=1  1是好的，全部都是好的，结果把2跳过了，好的中有一个坏的return  right=mid;}}//  返回最最左边好的return left;}}

LC 35 搜索插入位置

class Solution {public int searchInsert(int[] nums, int target) {int n = nums.length;int left = 0, right = n - 1, ans = n;while (left <= right) {int mid = ((right - left) >> 1) + left;if (target <= nums[mid]) {ans = mid;right = mid - 1;} else {left = mid + 1;}}return ans;}
}

LC997有序数组的平方

// 方法1最简单的方法就是将数组 nums 中的数平方后直接排序
//时间复杂度：O(nlogn)，其中 n 是数组 nums 的长度。
//空间复杂度：O(logn)。除了存储答案的数组以外，我们需要O(logn) 的栈空间进行排序。class Solution {public int[] sortedSquares(int[] nums) {int[] newArray=new int [nums.length];for(int i=0;i<nums.length;i++){newArray[i]=nums[i]*nums[i];} Arrays.sort(newArray);return newArray;}
}

@LC1295统计位数为偶数的数字22-10–27

class Solution {public int findNumbers(int[] nums) {int count=0;//1.遍历数组for(int i=0;i<nums.length;i++){//3.判断偶数if(fib( nums[i] ) %2 ==0){count++;}}return count;}//2.判断位数public int fib(int n){int count=0;while(n>0){n/=10;count++;}return count;}}class Solution {public int findNumbers(int[] nums) {int count=0;for(int i=0;i<nums.length;i++){//  valueof()类型转换String a=String.valueOf(nums[i]);if(a.length()%2==0){count++;}}return count;}
}

LC1832. 判断句子是否为全字母句

 public static boolean checkIfPangram(String sentence) {// 1.定义数组用于记录改变过的值，只要有26个字母，每一个都会改成-1，否则就是默认值0int[] count=new int[26];//2.把string sentence 变成字符数组，遍历拿到一个个元素for (char i:sentence.toCharArray()){//abc//97-97 =0   -1 //98-97=1   -1// 2拿到数组下标，将值改-1count[i-'a']--;}for (int i = 0; i < 26; i++) {//数组下标的值，等于0 说明没改if(count[i]==0){return false;}}//说明改了，有这个字母return true;}

LC283 移动0和LC 27 方法一样

这个录得有视频

   // 双指针方法public void moveZeroes(int[] nums) {//1.判断边界if(nums==null){return;}// j记录0的值 和交换数值int j=0;for (int i = 0; i < nums.length; i++) {if(nums[i]!=0){nums[j++]=nums[i];}}// j中0的位置 放到最后 5个元素，j=4 支循环2次for (int i = j; i <nums.length ; i++) {nums[i]=0;}}

LC27. 移除元素

给你一个数组 nums 和一个值 val，你需要原地移除所有数值等于 val 的元素，并返回移除后数组的新长度。

不要使用额外的数组空间，你必须仅使用 O(1) 额外空间并原地修改输入数组。

元素的顺序可以改变。你不需要考虑数组中超出新长度后面的元素。

public int removeElement(int[] nums, int val) {int j=0;for(int i=0;i<nums.length;i++){if(nums[i]!=val){nums[j++]=nums[i];}}return j;}

LC 344. 反转字符串双针针解法

  public void reverseString(char[] s) {int n=s.length;// a b c d ei		 ji  j//，假设反转前字符数组为 s[0] s[1] s[2] ... s[N - 1]，那么反转后字符数组为 s[N - 1] s[N - 2] ... s[0]。for(int i=0,j=n-1;i<j;i++,j--){char temp=s[i];s[i]=s[j];s[j]=temp;}}

LC 344. 字符串反转

  private static void fib4() {//练习4：字符串反转String str="abcdefg";//1.把字符串转成字符数组char[] ch = str.toCharArray();//2.借用辅助空间char[] char1= new char[ch.length];//3.数组反转for (int i = 0,j=ch.length-1; i <ch.length ; i++) {char1[j--]=ch[i];}//4字符数组转成字符串String str2 = new String(char1);System.out.println(str2);}

@10-20-LC80. 删除有序数组中的重复项 II

  public int removeDuplicates(int[] nums) {int j=2,i=2;for(i=2;i<nums.length;i++){//如果快指针和满指针-2下标处值不一样//1快指针的值赋给满指针，同时++// 2如果一样  满指针不懂，快指针移动 继续重复1if(nums[j-2]!=nums[i]){nums[j++]=nums[i];}}return j;
}

10-20-LC26删除有序数组中的重复项

   public int removeDuplicates(int[] nums) {int i=0;int j=0;for(i=0;i<nums.length;i++){// 如果i所在位置的值，不等于j所在位置的值if(nums[j]!=nums[i]){//2.j位置后面的值改成nums[i]的值//3.j向后移动继续比较，相同跳过，不相同继续nums[++j]=nums[i];}}return j+1;}

第二部分：哈希表

@1021—LC349. 两个数组的交集-set

给定两个数组 nums1 和 nums2 ，返回 它们的交集 。输出结果中的每个元素一定是唯一的。我们可以 不考虑输出结果的顺序 。

class Solution {public int[] intersection(int[] nums1, int[] nums2) {if(nums1==null || nums1.length==0 || nums2==null ||nums2.length==0){return new int[0];}Set <Integer> set1=new HashSet<>();Set <Integer>  result=new HashSet<>();//把数组1的元素放进去，自动去重for(int i:nums1){set1.add(i);}//把数组2和数组1重复的元素放到resultfor(int i:nums2){if(set1.contains(i)){result.add(i);}}// 集合转数组的方法1  慢很多//return result.stream().mapToInt(x->x).toArray();//定义一个数组，数组取遍历集合int[] arr=new int[result.size()];int index=0;for(int i:result){arr[index++]=i;}return arr;}
}

1021-LC242. 有效的字母异位词-哈希数组

class Solution {public boolean isAnagram(String s, String t) {//利用数组下标方法int hasp[]=new int[26];//  第一个数组有就加for(int i=0;i<s.length();i++){hasp[s.charAt(i)-'a']++;}//  第二个数组有就加for(int j=0;j<t.length();j++){hasp[t.charAt(j)-'a']--;}// 两个字符串相同，一加一减值=0for(int k=0;k<hasp.length;k++){if(hasp[k]!=0){return false;}}return true;}
}

1021-LC217存在重复元素-Set

class Solution {//  思路1：排序后排序前后是否重复/*public boolean containsDuplicate(int[] nums) {Arrays.sort(nums);for (int i = 0; i <nums.length-1; i++) {if(nums[i]==nums[i+1]){return true;}}return false;}*/public boolean containsDuplicate(int[] nums) {Set<Integer> set =new HashSet<>();for(int x:nums){if(!set.add(x)){return true;}}return false;}}

10-21-LC219存在重复元素2-滑动窗口

class Solution {public boolean containsNearbyDuplicate(int[] nums, int k) {Set<Integer> set =new HashSet<>();for(int i=0;i<nums.length;i++){//找到了重复的元素if(set.contains(nums[i])){return true;}//没找到就加入set中 扩大窗口set.add(nums[i]);//滑动窗口超过了指定大小，缩小窗口if(set.size()>k){set.remove(nums[i-k]);}}return false;}
}

@10-22-LC202快乐数-哈希

package day0;import java.util.HashSet;
import java.util.Set;/*** @Description TODO* @Author 李国勇* @Date 2022/10/22 22:46*/
public class LC205_快乐数 {public boolean isHappy(int n) {Set<Integer> set = new HashSet<>();while (n != 1 && !set.contains(n)) {set.add(n);n = getNextNumber(n);}return n == 1;}private int getNextNumber(int n) {int sum = 0;while (n > 0) {int t = n % 10;sum += t * t;n /= 10;}return sum;}
}

10-22-LC15三数之和+哈希去重

class Solution {public List<List<Integer>> threeSum(int[] nums) {List<List<Integer>> result=new ArrayList<>();Arrays.sort(nums);for(int i=0;i<nums.length;i++){// 如果三数之和>0,第一个数》0结束if(nums[i]>0){return result;}// 对a去重if(i>0 && nums[i]==nums[i-1]){continue;}int left=i+1;int right=nums.length-1;while(right>left){int sum=nums[i]+nums[left]+nums[right];if(sum>0){right--;}else if(sum<0){left++;}else{result.add(Arrays.asList(nums[i],nums[left],nums[right]));// 对b，c进行去重while(right>left && nums[right]==nums[right-1]){right--;}while(right>left && nums[left]==nums[left+1]){left++;}right--;left++;}}}return result;}
}

@10-23-LC645. 错误的集合

package day0;import java.util.Arrays;/*** @Description TODO* @Author 李国勇* @Date 2022/10/23 9:47*/
public class LC645_n数组重复一个元素 {//bug 前提必须从1开始，连续的，重复的重现2次// 数组记录重现出现2次和出现0次的public static void main(String[] args) {int nums[]={1,2,2};System.out.println(findErrorNums(nums));}public static String findErrorNums(int[] nums) {int n=nums.length;int[] copy=new int[n+1];int[] ans=new int[2];for(int i=0;i<nums.length;i++){copy[nums[i]]++;}for(int i=1;i<=n;i++){if(copy[i]==0) {ans[1]=i;}if(copy[i]==2) {ans[0]=i;}}return Arrays.toString(ans);}
}

10-23-LC1768交替合并字符串

class Solution {public String mergeAlternately(String word1, String word2) {int m=word1.length();int n=word2.length();int i=0,j=0;StringBuilder ans=new StringBuilder();while(i<m||j<n){if(i<m){ans.append(word1.charAt(i));i++;}if(j<n){ans.append(word2.charAt(j));j++;}}return ans.toString();}
}

10-23-LC454四数相加2(四个数组)

class Solution {public int fourSumCount(int[] nums1, int[] nums2, int[] nums3, int[] nums4) {Map<Integer, Integer> map = new HashMap<>();int temp;int res = 0;//统计两个数组中的元素之和，同时统计出现的次数，放入mapfor (int i : nums1) {for (int j : nums2) {temp = i + j;if (map.containsKey(temp)) {map.put(temp, map.get(temp) + 1);} else {map.put(temp, 1);}}}//统计剩余的两个元素的和，在map中找是否存在相加为0的情况，同时记录次数for (int i : nums3) {for (int j : nums4) {temp = i + j;if (map.containsKey(0 - temp)) {res += map.get(0 - temp);}}}return res;}
}

10-23-LC18四数之和

package day0;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;/*** @Description TODO* @Author 李国勇* @Date 2022/10/23 16:23*/
public class LC18四数之和 {public List<List<Integer>> fourSum(int[] nums, int target) {List<List<Integer>> result = new ArrayList<>();Arrays.sort(nums);for (int i = 0; i < nums.length; i++) {// nums[i] > target 直接返回, 第一次剪枝操作if (nums[i] > 0 && nums[i] > target) {return result;}//第一次去重操作if (i > 0 && nums[i - 1] == nums[i]) {continue;}for (int j = i + 1; j < nums.length; j++) {// 第二次剪枝去重if (j > i + 1 && nums[j - 1] == nums[j]) {continue;}int left = j + 1;int right = nums.length - 1;while (right > left) {long sum = (long) nums[i] + nums[j] + nums[left] + nums[right];if (sum > target) {right--;} else if (sum < target) {left++;} else {result.add(Arrays.asList(nums[i], nums[j], nums[left], nums[right]));while (right > left && nums[right] == nums[right - 1]) {right--;}while (right > left && nums[left] == nums[left + 1]) {left++;}left++;right--;}}}}return result;}
}

10-23-LC18四数之和-官方题解

class Solution {public List<List<Integer>> fourSum(int[] nums, int target) {List<List<Integer>> quadruplets = new ArrayList<List<Integer>>();if (nums == null || nums.length < 4) {return quadruplets;}Arrays.sort(nums);int length = nums.length;for (int i = 0; i < length - 3; i++) {if (i > 0 && nums[i] == nums[i - 1]) {continue;}if ((long) nums[i] + nums[i + 1] + nums[i + 2] + nums[i + 3] > target) {break;}if ((long) nums[i] + nums[length - 3] + nums[length - 2] + nums[length - 1] < target) {continue;}for (int j = i + 1; j < length - 2; j++) {if (j > i + 1 && nums[j] == nums[j - 1]) {continue;}if ((long) nums[i] + nums[j] + nums[j + 1] + nums[j + 2] > target) {break;}if ((long) nums[i] + nums[j] + nums[length - 2] + nums[length - 1] < target) {continue;}int left = j + 1, right = length - 1;while (left < right) {long sum = (long) nums[i] + nums[j] + nums[left] + nums[right];if (sum == target) {quadruplets.add(Arrays.asList(nums[i], nums[j], nums[left], nums[right]));while (left < right && nums[left] == nums[left + 1]) {left++;}left++;while (left < right && nums[right] == nums[right - 1]) {right--;}right--;} else if (sum < target) {left++;} else {right--;}}}}return quadruplets;}
}

第三部分：字符串

10-23-LC344. 反转字符串

class Solution {public static void reverseString(char[] s) {int len=s.length;for(int i=0,j=s.length-1;i<len/2;i++,j--){char temp=s[i];s[i]=s[j];s[j]=temp;}}
}

10-23-LC541. 反转字符串2–有三种解法

package day0;/*** @Description TODO* @Author 李国勇* @Date 2022/10/23 19:47*/
public class LC541_反转字符串2 {public static void main(String[] args) {String s="abcdefxy";System.out.println(reverseStr(s, 3));}public static String reverseStr(String s, int k) {char[] ch = s.toCharArray();// 1. 每隔 2k 个字符的前 k 个字符进行反转===6个for (int i = 0; i < ch.length; i = i + 2 * k) {// 2. if剩余字符小于 2k 但大于或等于 k 个，则反转前 k 个字符===3个if (i + k <= ch.length) {reverse(ch, i, i + k - 1);continue;}// 2. else剩余字符少于 k 个，则将剩余字符全部反转  只剩2个reverse(ch, i, ch.length - 1);}return new String(ch);}private static void reverse(char[] ch, int i, int j) {for(; i<j;i++,j--){char temp=ch[i];ch[i]=ch[j];ch[j]=temp;}}
}

10-23-LC151. 反转字符串中的单词

package day0;/*** @Description TODO* @Author 李国勇* @Date 2022/10/23 22:13*/
public class LC151_反转字符串中的单词 {public static void main(String[] args) {System.out.println(reverseWords("  he wo "));}public static String reverseWords(String s) {// 1.去除首尾以及中间多余空格StringBuilder sb = removeSpace(s);// 2.反转整个字符串reverseString(sb, 0, sb.length() - 1);// 3.反转各个单词reverseEachWord(sb);return sb.toString();}private static void reverseEachWord(StringBuilder sb) {int start = 0;int end = 1;int n = sb.length() - 1;while (start < n) {while (end < n && sb.charAt(end) != ' ') {end++;}reverseString(sb, start, end - 1);start = end + 1;end = start + 1;}}// 2.反转整个字符串private static void reverseString(StringBuilder sb, int start, int end) {while (start < end) {char temp = sb.charAt(start);sb.setCharAt(start, sb.charAt(end));sb.setCharAt(end, temp);start++;end--;}}private static StringBuilder removeSpace(String s) {int start = 0;int end = s.length() - 1;while (s.charAt(start) == ' ') {start++;}while (s.charAt(end) == ' ') {end--;}StringBuilder sb = new StringBuilder();while (start <= end) {char c = s.charAt(start);if (c != ' ' || sb.length() - 1 != ' ') {sb.append(c);}start++;}return sb;}
}

@10-25-LC1662. 检查两个字符串数组是否相等

   public boolean arrayStringsAreEqual(String[] word1, String[] word2) {StringBuilder s1=new StringBuilder();StringBuilder s2=new StringBuilder();for (int i = 0; i < word1.length; i++) {s1.append(word1[i]);}for (int i = 0; i < word2.length; i++) {s2.append(word2[i]);}return s1.toString().equals(s2.toString());}

@10-27-LC1822元素累积的符号

class Solution {public int arraySign(int[] nums) {int flag=1;for(int i:nums){if(i==0){return 0;}if(i<0){flag=-flag;}}return flag;}
}

10-29-LC200-DFS

class Solution {public void dfs(char[][] grid,int i,int j){int row=grid.length;int col=grid[0].length;if(i<0 || j<0 || i >=row || j >=col ||grid[i][j]=='0'){return;}grid[i][j]='0';dfs(grid,i-1,j);dfs(grid,i+1,j);dfs(grid,i,j-1);dfs(grid,i,j+1);}public int numIslands(char[][] grid) {if(grid == null && grid.length==0){return 0;}int row=grid.length;int col=grid[0].length;int result=0;for(int i=0;i<row;i++){for(int j=0;j<col;j++){if(grid[i][j]=='1'){result++;dfs(grid,i,j);}}}return result;}}

i++){
for(int j=0;j<col;j++){
if(grid[i][j]==‘1’){
result++;
dfs(grid,i,j);
}
}
}
return result;
}

}

LeetCode笔记