leetcode First Missing Positive 首个缺失的正数，交换系列

1.leetcode First Missing Positive 首个缺失的正数

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.

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]);
            }
        }
        for (int i = 0; i < n; ++i) {
            if (nums[i] != i + 1) return i + 1;
        }
        return n + 1;
    }
};

2.Missing Number 丢失的数字

Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.

For example,
Given nums = [0, 1, 3] return 2.

Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?

这道题还可以用二分查找法来做，我们首先要对数组排序，然后我们用二分查找法算出中间元素的下标，然后用元素值和下标值之间做对比，如果元素值大于下标值，则说明缺失的数字在左边，此时将right赋为mid，反之则将left赋为mid+1。那么看到这里，作为读者的你可能会提出，排序的时间复杂度都不止O(n)，何必要多此一举用二分查找，还不如用上面两种方法呢。对，你说的没错，但是在面试的时候，有可能人家给你的数组就是排好序的，那么此时用二分查找法肯定要优于上面两种方法，

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int left = 0, right = nums.size();
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] > mid) right = mid;
            else left = mid + 1;
        }
        return right;
    }
};