题目意思是找出第一个没出现的最小正整数。

Explanation:

1. Move Numbers to Correct Positions:

   • The idea is to place each number in its corresponding index. For example, 1 should be at index 0, 2 should be at index 1, and so on. This is done using a while loop to swap the elements until all valid numbers (those within the range 1 to n) are in their respective positions.

2. Find the First Missing Positive:

   • After the first step, iterate through the array. The first position where the element is not equal to i + 1 is the smallest missing positive integer.

3. Return n + 1 if all numbers are in correct places:

   • If all numbers are in their correct places, the smallest missing positive integer must be n + 1.

Time Complexity:

• The time complexity is $O(n)$ because each element is swapped at most once.

Space Complexity:

• The space complexity is $O(1)$ since we are using constant extra space.

class Solution {
public:int firstMissingPositive(vector<int>& nums) {int n = nums.size(); //由于数组下标是从0开始，所以值为 n 的元素的下标按顺序是 n - 1//例如1的下标是0，2的下标是1...for(int i = 0; i < n; i++) {//仅当当前元素值的范围介于[1,n]之间，并且它所该移动到的位置上的元素不相等时我们移动它到这个位置//值为 nums[i] 的元素按顺序排列的下标是 nums[i] - 1while(nums[i] > 0 && nums[i] <=n && nums[i] != nums[nums[i] - 1]) {swap(nums[i], nums[nums[i] - 1]);}}//然后，我们遍历数组，当下标 i 对应的值不等于 i + 1 时，返回 i + 1for(int i = 0; i < n; i++) {if(nums[i] != i + 1) {return i + 1;}}//如果在上面的循环中并没有return，说明数组所以元素值被排列成了 1,2,...,nreturn n + 1;}
};