回溯Backtracking Algorithm

devtools/2025/1/18 7:00:59/

目录

1) 入门例子

2) 全排列-Leetcode 46

3) 全排列II-Leetcode 47

4) 组合-Leetcode 77

5) 组合总和-Leetcode 39

6) 组合总和 II-Leetcode 40

7) 组合总和 III-Leetcode 216

8) N 皇后 Leetcode 51

9) 解数独-Leetcode37

10) 黄金矿工-Leetcode1219

其它题目

1) 入门例子

javascript">/*** 回溯* * 程序在运行过程中分成了多个阶段* 通过某些手段,将数据恢复到之前某一阶段,这就称之为回溯* 手段包括*      方法栈*      自定义栈*/
public class Backtracking {public static void main(String[] args) {rec(1);}private static void rec(int n) {if(n==3){return ;}System.out.println(n);rec(n+1);System.out.println(n);}
}

 

 方法栈

如果是集合又有什么样的效果呢?

如果用的是可变的集合或者数组必须手动的恢复集合状态
java">public class Backtracking {public static void main(String[] args) {rec(1, new LinkedList<>());}static void rec(int n, LinkedList<String> list) {if (n == 3) {return;}System.out.println("before:" + list);list.push("a");rec(n + 1, list);list.pop();System.out.println("after:" + list);}
}

2) 全排列-Leetcode 46

java">public class PermuteLeetcode46 {static List<List<Integer>> permute(int[] nums) {boolean[] visited = new boolean[nums.length];LinkedList<Integer> stack = new LinkedList<>();List<List<Integer>> r = new ArrayList<>();rec(nums, visited, stack, r);return r;}static void rec(int[] nums, boolean[] visited, LinkedList<Integer> stack, List<List<Integer>> r) {if (stack.size() == nums.length) {r.add(new ArrayList<>(stack));return;}for (int i = 0; i < nums.length; i++) {if(visited[i]){continue;}stack.push(nums[i]);visited[i] = true;rec(nums, visited, stack, r);stack.pop();visited[i] = false;}}public static void main(String[] args) {List<List<Integer>> permute = permute(new int[]{1, 2, 3});for (List<Integer> s : permute) {System.out.println(s);}}
}

3) 全排列II-Leetcode 47

47. 全排列 II - 力扣（LeetCode）

java">public class PermuteLeetcode47 {static List<List<Integer>> permuteUnique(int[] nums) {Arrays.sort(nums);List<List<Integer>> result = new ArrayList<>();dfs(nums, new boolean[nums.length], new LinkedList<>(), result);return result;}static void dfs(int[] nums, boolean[] visited, LinkedList<Integer> stack, List<List<Integer>> result) {if (stack.size() == nums.length) {result.add(new ArrayList<>(stack));return;}for (int i = 0; i < nums.length; i++) {if (i > 0 && nums[i] == nums[i - 1] && !visited[i-1]) { // 找出重复数字continue;}if (!visited[i]) {stack.push(nums[i]);visited[i] = true;dfs(nums, visited, stack, result);visited[i] = false;stack.pop();}}}public static void main(String[] args) {int[] nums = {1, 1, 3};        List<List<Integer>> permute = permuteUnique(nums);for (List<Integer> list : permute) {System.out.println(list);}}
}

  • 排好序，这样重复的数字会相邻

  • 定好规则：必须 1 固定之后才能固定 1'，即 1 的 visited = true 才能继续处理 1'

  • 在遍历时，遇到了 nums[i] == nums[i - 1]（即 1 和 1‘ 这种情况），进一步检查 i-1 位置的数字有没有 visited，没有，则不处理（剪枝）

4) 组合-Leetcode 77

k=2 

k=3 

 

java">class Solution {public List<List<Integer>> combine(int n, int k) {List<List<Integer>>result = new ArrayList<>();LinkedList<Integer>stack = new LinkedList<>();dfs(1,n,k,stack,result);return result;}public void dfs(int start,int n,int k,LinkedList<Integer>stack,List<List<Integer>>result){if(stack.size()==k){result.add(new ArrayList<>(stack));return;}for(int i = start;i<=n;i++){stack.push(i);dfs(i+1,n,k,stack,result);stack.pop();}}
}

 

减枝

k- stack.length  还差几个能凑满

n - i +1 还剩下几个备用数字   if(k-stack.length >n-i+1) continue;

java">class Solution {public List<List<Integer>> combine(int n, int k) {List<List<Integer>>result = new ArrayList<>();LinkedList<Integer>stack = new LinkedList<>();dfs(1,n,k,stack,result);return result;}public void dfs(int start,int n,int k,LinkedList<Integer>stack,List<List<Integer>>result){if(stack.size()==k){result.add(new ArrayList<>(stack));return;}for(int i = start;i<=n;i++){if(k - stack.size() > n-i+1){continue;}stack.push(i);dfs(i+1,n,k,stack,result);stack.pop();}}
}

 

java">public class CombinationLeetcode77 {static List<List<Integer>> combinationSum(int n, int k) {List<List<Integer>> result = new ArrayList<>();dfs(n, k, 1, new LinkedList<>(), result);return result;}static int count = 0;static void dfs(int n, int k, int start, LinkedList<Integer> stack, List<List<Integer>> result) {count++;if (k == 0) {result.add(new ArrayList<>(stack));System.out.println(stack);return;}
//      if (k > n - start + 1) { return; }for (int i = start; i <= n; i++) {
//            System.out.printf("k-1=%d n=%d i=%d %n", k - 1, n, i);if (k > n - i + 1) {continue;}stack.push(i);dfs(n, k - 1, i + 1, stack, result);stack.pop();}}public static void main(String[] args) {List<List<Integer>> lists = combinationSum(5, 4);
//        for (List<Integer> list : lists) {
//            System.out.println(list);
//        }System.out.println(count);}
}

 

  • k 代表剩余要组合的个数

  • n - i + 1 代表剩余可用数字

  • 剪枝条件是：剩余可用数字要大于剩余组合数

  • 为啥放在外面不行？即这行代码：if (k > n - start + 1) { return; }

    • 因为它只考虑了 start 一种情况，而实际在循环内要处理的是 start，start+1 ... n 这多种情况

似乎 ArrayList 作为 stack 性能高一些，见下面代码，但是这道题在 leetcode 上执行时间不稳定，相同代码都会有较大时间差异（15ms vs 9ms）

java">class Solution {public List<List<Integer>> combine(int n, int k) {        List<List<Integer>> result = new ArrayList<>();if(k == 0 || n < k) return result;dfs(n, k, 1, new ArrayList<>(), result);return result;}static void dfs(int n, int k, int start, ArrayList<Integer> stack, List<List<Integer>> result) {if (k == 0) {result.add(new ArrayList<>(stack));return;}for (int i = start; i <= n; i++) {if (k-1 > n - i) {continue;}stack.add(i);dfs(n, k - 1, i + 1, stack, result);stack.remove(stack.size()-1);}}
}

5) 组合总和-Leetcode 39

java">class Solution {public List<List<Integer>> combinationSum(int[] candidates, int target) {List<List<Integer>>result = new ArrayList<>();LinkedList<Integer> stack= new LinkedList<>();dfs(0,candidates,target,stack,result);return result;}public void dfs(int start,int[] candidates,int target,LinkedList<Integer>stack,List<List<Integer>>result){if(target<0)return ;if(target==0){result.add(new ArrayList<>(stack));return;}for(int i = start;i<candidates.length;i++){int candidate = candidates[i];stack.push(candidate);dfs(i,candidates,target-candidate,stack,result);stack.pop();}}
}

 

java">class Solution {public List<List<Integer>> combinationSum(int[] candidates, int target) {List<List<Integer>>result = new ArrayList<>();LinkedList<Integer> stack= new LinkedList<>();dfs(0,candidates,target,stack,result);return result;}public void dfs(int start,int[] candidates,int target,LinkedList<Integer>stack,List<List<Integer>>result){if(target==0){result.add(new ArrayList<>(stack));return;}for(int i = start;i<candidates.length;i++){int candidate = candidates[i];if(target<candidate){continue;}stack.push(candidate);dfs(i,candidates,target-candidate,stack,result);stack.pop();}}
}

与之前的零钱兑换问题其实是一样的，只是

  • 本题求的是：所有组合的具体信息

  • 零钱兑换问题求的是：所有组合中数字最少的、所有组合个数... [动态规划]

6) 组合总和 II-Leetcode 40

java">public class CombinationLeetcode40 {static List<List<Integer>> combinationSum2(int[] candidates, int target) {Arrays.sort(candidates);List<List<Integer>> result = new ArrayList<>();dfs(target, 0, candidates, new boolean[candidates.length], new LinkedList<>(), result);return result;}static void dfs(int target, int start, int[] candidates, boolean[] visited, LinkedList<Integer> stack, List<List<Integer>> result) {if (target == 0) {result.add(new ArrayList<>(stack));return;}for (int i = start; i < candidates.length; i++) {int candidate = candidates[i];if (target < candidate) {continue;}if (i > 0 && candidate == candidates[i - 1] && !visited[i - 1]) {continue;}visited[i] = true;stack.push(candidate);dfs(target - candidate, i + 1, candidates, visited, stack, result);stack.pop();visited[i] = false;}}public static void main(String[] args) {int[] candidates = {10, 1, 2, 7, 6, 1, 5};        List<List<Integer>> lists = combinationSum2(candidates, 8);for (List<Integer> list : lists) {System.out.println(list);}}
}

7) 组合总和 III-Leetcode 216

java">class Solution {public List<List<Integer>> combinationSum3(int k, int target ) {List<List<Integer>>result = new ArrayList<>();dfs(1,target,k,new ArrayList<>(),result);return result;}//static int count = 0;static void dfs(int start,int target,int k,ArrayList<Integer>stack,List<List<Integer>>result){// count++;if(target==0&&stack.size()==k){result.add(new ArrayList<>(stack));return;}for(int i = start;i<=9;i++){// 还差几个数字       剩余可用数字//  if(k-stack.size() > 9-i+1){//    continue;//} 这个减枝效率较低 设置一个count变量即可查看if(target<i){continue;}if(stack.size()==k){continue;}stack.addLast(i);dfs(i+1,target-i,k,stack,result);stack.removeLast();}}
}

8) N 皇后 Leetcode 51

左斜线处理 i+j 相等i - j

 n-1-(i-j)

 i== n 找到解

java">public class NQueenLeetcode51 {static List<List<String>> solveNQueens(int n) {List<List<String>> result = new ArrayList<>();char[][] table = new char[n][n];//'.' 'Q'boolean[] va = new boolean[n];//列冲突boolean[] vb = new boolean[2 * n - 1];//左斜线冲突boolean[] vc = new boolean[2 * n - 1];//右斜线冲突for (int i = 0; i < n; i++) {Arrays.fill(table[i], '.');}dfs(0, n, table, result, va, vb, vc);return result;}static void dfs(int i, int n, char[][] table, List<List<String>> result, boolean[] va, boolean[] vb, boolean[] vc) {if (i == n) {ArrayList<String> list = new ArrayList<>();for (char[] chars : table) {list.add(String.valueOf(chars));}result.add(list);return;}for (int j = 0; j < n; j++) {if (va[j] || vb[i + j] || vc[n - 1-(i-j)]) {continue;}va[j] = true;vb[i + j] = true;vc[n-1-(i-j)] = true;table[i][j] = 'Q';dfs(i + 1, n, table, result, va, vb, vc);table[i][j] = '.';va[j] = false;vb[i + j] = false;vc[i - j + n - 1] = false;}}public static void main(String[] args) {int count = 0;for (List<String> table : solveNQueens(4)) {for (String row : table) {System.out.println(row);}count++;System.out.println("--------------------- " + count);}}
}
java">public class NQueenLeetcode51 {static List<List<String>> solveNQueens(int n) {List<List<String>> result = new ArrayList<>();char[][] table = new char[n][n];for (int i = 0; i < n; i++) {Arrays.fill(table[i], '.');}dfs(0, n, table, result);return result;}static void dfs(int i, int n, char[][] table, List<List<String>> result) {if (i == n) {ArrayList<String> list = new ArrayList<>();for (char[] chars : table) {list.add(String.valueOf(chars));}result.add(list);return;}for (int j = 0; j < n; j++) {if (notValid(table, i, j)) {continue;}table[i][j] = 'Q';dfs(i + 1, n, table, result);table[i][j] = '.';}}/*.   .   .   ..   .   .   ..   ?   .   ..   .   .   .*/static boolean notValid(char[][] table, int row, int col) {int n = table.length;for (int i = 0; i < n; i++) {if (table[i][col] == 'Q') { // 上return true;}}for (int i = row - 1, j = col - 1; i >= 0 && j >= 0; i--, j--) {if (table[i][j] == 'Q') {return true;}}for (int i = row - 1, j = col + 1; i >= 0 && j < n; i--, j++) {if (table[i][j] == 'Q') {return true;}}return false;}public static void main(String[] args) {int count = 0;for (List<String> table : solveNQueens(8)) {for (String row : table) {System.out.println(row);}count++;System.out.println("--------------------- " + count);}}
}

9) 解数独-Leetcode37

判断在那个九宫格 ==>  i/3*3+j/3  

java"> public void solveSudoku(char[][] board) {/*1.不断遍历每个未填的空格逐一尝试1~9 若行,列,九宫格内没有冲突,则填入2.一旦1~9 都尝试失败,回溯到上一次状态,换数字填入3.关键还是要记录冲突状态*/// 行冲突状态boolean[][] ca =new boolean[9][9];// ca[i] = {false,false,true,true,true,true...}// 列冲突状态boolean[][] cb = new boolean[9][9];// cb[j] = {false,true,true....}// 九宫格冲突状态//i/3*3+j/3 = ..在几号九宫格boolean[][] cc = new boolean[9][9];//cc[i/3*3+j/3] = {...}for(int i =0;i<9;i++){for(int j=0;j<9;j++){char ch = table[i][j];if(ch!='.'){//初始化冲突状态ca[i][ch-'1']=true;     //'5'- '1' --> 4cb[j][ch-'1']=true;cc[i/3*3+j/3][ch-'1'] =true;}}}dfs(0,0,table,ca,cb,cc);}static boolean dfs(int i,int j,char[][] table,boolean[][] ca,boolean[][] cb,boolean[][] cc){while(table[i][j]!='.'){ //查找下一个空格if(++j>=9){j=0;i++;//到下一行}if(i>=9){return true; //找到解了}}//填空for(int x = 1;x<=9;x++){//检查冲突if(ca[i][x-1]||cb[j][x-1]||cc[i/3*3+j/3][x-1]){continue;}table[i][j] =(char)x+'0';   //1 -> '1'//ca[0][0] = true;  第0行不能存储'1'//cb[2][0] = true;  第2列不能存储'1'//cc[0][0] = true;  第0个九宫格不能存储'1'ca[i][x-1] = true;cb[j][x-1] = true;cc[i/3*3+j/3][x-1]=true;if(dfs(i,j,table,ca,cb,cc)){return true;}table[i][j] = '.';ca[i][x-1] = false;cb[j][x-1] = false;cc[i/3*3+j/3][x-1]=false;}return false;}
java">public class SudokuLeetcode37 {static void solveSudoku(char[][] table) {int n = 9;boolean[][] va = new boolean[n][n];//行冲突boolean[][] vb = new boolean[n][n];//列冲突boolean[][][] vc = new boolean[3][3][n];//九宫格冲突for (int i = 0; i < n; i++) {for (int j = 0; j < n; j++) {if (table[i][j] != '.') {int x = table[i][j] - '0' - 1;va[i][x] = true;vb[j][x] = true;vc[i / 3][j / 3][x] = true;}}}dfs(table, va, vb, vc, 0, 0);}static boolean dfs(char[][] table, boolean[][] va, boolean[][] vb, boolean[][][] vc, int i, int j) {while (table[i][j] != '.') {if (++j >= 9) {j = 0;i++;}if (i >= 9) {return true;}}int n = table.length;for (int d = 0; d < n; d++) {if (va[i][d] || vb[j][d] || vc[i / 3][j / 3][d]) {continue;}char ch = (char) (d + '0' + 1);table[i][j] = ch;va[i][d] = true;vb[j][d] = true;vc[i / 3][j / 3][d] = true;boolean dfs = dfs(table, va, vb, vc, i, j);if (dfs) {return true;}table[i][j] = '.';va[i][d] = false;vb[j][d] = false;vc[i / 3][j / 3][d] = false;}return false;}public static void main(String[] args) {char[][] table = {{'5', '3', '.', '.', '7', '.', '.', '.', '.'},{'6', '.', '.', '1', '9', '5', '.', '.', '.'},{'.', '9', '8', '.', '.', '.', '.', '6', '.'},{'8', '.', '.', '.', '6', '.', '.', '.', '3'},{'4', '.', '.', '8', '.', '3', '.', '.', '1'},{'7', '.', '.', '.', '2', '.', '.', '.', '6'},{'.', '6', '.', '.', '.', '.', '2', '8', '.'},{'.', '.', '.', '4', '1', '9', '.', '.', '5'},{'.', '.', '.', '.', '8', '.', '.', '7', '9'}};solveSudoku(table);print(table);}static char[][] solved = {{'5', '3', '4', '6', '7', '8', '9', '1', '2'},{'6', '7', '2', '1', '9', '5', '3', '4', '8'},{'1', '9', '8', '3', '4', '2', '5', '6', '7'},{'8', '5', '9', '7', '6', '1', '4', '2', '3'},{'4', '2', '6', '8', '5', '3', '7', '9', '1'},{'7', '1', '3', '9', '2', '4', '8', '5', '6'},{'9', '6', '1', '5', '3', '7', '2', '8', '4'},{'2', '8', '7', '4', '1', '9', '6', '3', '5'},{'3', '4', '5', '2', '8', '6', '1', '7', '9'}};static void print(char[][] table) {for (char[] chars : table) {System.out.println(new String(chars));}System.out.println(Arrays.deepEquals(table, solved));}
}
java">public class SudokuLeetcode37 {record Pair(int i, int j) {}static void solveSudoku(char[][] table) {int n = 9;boolean[][] va = new boolean[n][n];boolean[][] vb = new boolean[n][n];boolean[][][] vc = new boolean[3][3][n];List<Pair> blanks = new ArrayList<>();for (int i = 0; i < n; i++) {for (int j = 0; j < n; j++) {if (table[i][j] != '.') {int x = table[i][j] - '0' - 1;va[i][x] = true;vb[j][x] = true;vc[i / 3][j / 3][x] = true;} else {blanks.add(new Pair(i, j));}}}dfs(0, blanks, table, va, vb, vc);}static boolean dfs(int p, List<Pair> blanks, char[][] table, boolean[][] va, boolean[][] vb, boolean[][][] vc) {if (p == blanks.size()) {print(table);return true;}int n = table.length;for (int d = 0; d < n; d++) {Pair pair = blanks.get(p);if (va[pair.i][d] || vb[pair.j][d] || vc[pair.i / 3][pair.j / 3][d]) {continue;}char ch = (char) (d + '0' + 1);table[pair.i][pair.j] = ch;va[pair.i][d] = true;vb[pair.j][d] = true;vc[pair.i / 3][pair.j / 3][d] = true;boolean dfs = dfs(p + 1, blanks, table, va, vb, vc);if (dfs) {return true;}table[pair.i][pair.j] = '.';va[pair.i][d] = false;vb[pair.j][d] = false;vc[pair.i / 3][pair.j / 3][d] = false;}return false;}public static void main(String[] args) {char[][] table = {{'5', '3', '.', '.', '7', '.', '.', '.', '.'},{'6', '.', '.', '1', '9', '5', '.', '.', '.'},{'.', '9', '8', '.', '.', '.', '.', '6', '.'},{'8', '.', '.', '.', '6', '.', '.', '.', '3'},{'4', '.', '.', '8', '.', '3', '.', '.', '1'},{'7', '.', '.', '.', '2', '.', '.', '.', '6'},{'.', '6', '.', '.', '.', '.', '2', '8', '.'},{'.', '.', '.', '4', '1', '9', '.', '.', '5'},{'.', '.', '.', '.', '8', '.', '.', '7', '9'}};solveSudoku(table);print(table);}static char[][] solved = {{'5', '3', '4', '6', '7', '8', '9', '1', '2'},{'6', '7', '2', '1', '9', '5', '3', '4', '8'},{'1', '9', '8', '3', '4', '2', '5', '6', '7'},{'8', '5', '9', '7', '6', '1', '4', '2', '3'},{'4', '2', '6', '8', '5', '3', '7', '9', '1'},{'7', '1', '3', '9', '2', '4', '8', '5', '6'},{'9', '6', '1', '5', '3', '7', '2', '8', '4'},{'2', '8', '7', '4', '1', '9', '6', '3', '5'},{'3', '4', '5', '2', '8', '6', '1', '7', '9'}};static void print(char[][] table) {for (char[] chars : table) {System.out.println(new String(chars));}System.out.println(Arrays.deepEquals(table, solved));}
}

10) 黄金矿工-Leetcode1219

1219. 黄金矿工 - 力扣（LeetCode）

java">class Solution {int[][] g;boolean[][] vis;int m,n;int[][] dirs = new int[][]{{1,0},{-1,0},{0,1},{0,-1}};public int getMaximumGold(int[][] grid) {g = grid;m = g.length;n = g[0].length;vis= new boolean[m][n];int ans = 0;for(int i = 0;i<m;i++){for(int j = 0;j<n;j++){if(g[i][j]!=0){vis[i][j] = true;ans = Math.max(ans,dfs(i,j));vis[i][j] = false;}}}return ans;}int dfs(int x,int y){int ans = g[x][y];for(int[] d:dirs){int nx = x+d[0],ny = y+d[1];if(nx<0||nx>=m||ny<0||ny>=n)continue;if(g[nx][ny]==0)continue;if(vis[nx][ny]) continue;vis[nx][ny] = true;ans = Math.max(ans,g[x][y] + dfs(nx,ny));vis[nx][ny] =false;}return ans;}
}

其它题目

题号标题说明
Leetcode 1219黄金矿工
无马踏棋盘（The Knight’s tour problem）
无Rat in a Maze与 Leetcode 62 不同路径区别在于，该题问的是有多少种走法，而本题只是找到其中一种走法实现

543. 二叉树的直径 - 力扣（LeetCode）

java">class Solution {int ans = 0;public int diameterOfBinaryTree(TreeNode root) {dfs(root);return ans;}int dfs(TreeNode u){if(u==null) return 0;int l = dfs(u.left),r=dfs(u.right);ans = Math.max(ans,l+r);
//返回最大深度return Math.max(l,r)+1;}
}

433. 最小基因变化 - 力扣（LeetCode）

java">class Solution {int ans = Integer.MAX_VALUE;public int minMutation(String start, String end, String[] bank) {backtrack(start, end, bank, new boolean[bank.length], 0);return ans == Integer.MAX_VALUE ? -1 : ans;}public void backtrack(String start, String end, String[] bank, boolean[] used, int t) {if (t >= ans) return;if (start.equals(end)) {ans = Math.min(ans, t);} else {for (int i = 0, diff = 0; i < bank.length; i++, diff = 0) {if (used[i]) continue;for (int j = 0; j < start.length(); j++) diff += start.charAt(j) != bank[i].charAt(j) ? 1 : 0;if (diff == 1) {used[i] = true;backtrack(bank[i], end, bank, used, t + 1);used[i] = false;}}}}
}


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