41. First Missing Positive

description:

Given an unsorted integer array, find the smallest missing positive integer.
Note:
Your algorithm should run in O(n) time and uses constant extra space.
Example:

Example 1:

Input: [1,2,0]
Output: 3

Example 2:

Input: [3,4,-1,1]
Output: 2

Example 3:

Input: [7,8,9,11,12]
Output: 1

answer:

class Solution {
public:
    int firstMissingPositive(vector<int>& nums) {
        int n = nums.size();
        for(int i = 0; i < n; i++) {
            while (nums[i] > 0 && nums[i] <= n && nums[nums[i] - 1] != nums[i]) {
                swap (nums[i], nums[nums[i] - 1]); //先把数据放到正确的位置，就是说第一个位置放1，第二个位置放2
            }
        }
        
        for (int i = 0; i < n; i++) { // 然后进行遍历，那个位置不是应该的数就是答案
            if (nums[i] != i + 1) {
                return i + 1;
            }
        }
        return n + 1;
    }
};

relative point get√：

举个栗子：
3,4,-1,1

-1,4,3,1

-1,1,3,4

1,-1,3,4

hint :