Given an unsorted integer array, find the smallest missing positive integer.

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

Note:

Your algorithm should run in O(n) time and uses constant extra space.

解法
一开始的解法是开一个大小为n+1的数组，然后往数组里面塞数字，然后在遍历一遍数组得到结果。这种做法空间复杂度是O(n)。

class Solution {
public:
    int firstMissingPositive(vector<int>& nums) {
        int n = nums.size();
        vector<bool> arr(n+2, 0);
        for(auto num: nums) {
            if(num>=1&&num<=n+1)
                arr[num]=1;
        }
        for(int i=1;i<=n+1;i++) {
            if(arr[i]==false) 
                return i;
        }
    }
};

优化下，将操作全都放在nums里面进行，空间复杂度为O(1)

class Solution {
public:
    int firstMissingPositive(vector<int>& nums) {
        int n = nums.size();
        for(int i=0;i<nums.size();i++) {
            int num = nums[i];
            if(num>=1&&num<=n&&nums[i]!=nums[num-1]) {
                swap(nums[i], nums[num-1]);
                i--;
            }
        }
        for(int i=1;i<=n;i++) {
            if(nums[i-1]!=i)
                return i;
        }
        return n+1;
    }
};