Given an unsorted integer array, find the first missing positive integer.
For example,
Given [1,2,0] return 3,
and [3,4,-1,1] return 2.
Your algorithm should run in O(n) time and uses constant space.
题解：这道题目比较简单，需要注意的是几个陷阱：数组为空，数组里数据重复。
比较有技巧的地方是O(1)space，因为考虑到数据重复，此处只能排序， 这就要求重用nums数组空间。
引用网上一位网友的题解：
Put each number in its right place. For example: When we find 5, then swap it with A[4]. At last, the first place where its number is not right, return the place + 1.
程序：
1 class Solution { 2 public: 3 int firstMissingPositive(vector<int>& nums) { 4 for (int i = 0; i < nums.size(); i++) 5 { 6 while (nums[i] > 0 && nums[i] <= nums.size() && nums[i] != nums[nums[i] -1]) 7 { 8 swap(nums[i], nums[nums[i] -1]); 9 } 10 } 11 for (int i = 0; i < nums.size(); i++) 12 { 13 if (nums[i] != i + 1) 14 { 15 return i + 1; 16 } 17 } 18 return nums.size() + 1; 19 } 20 };
本文转自 jiu~ 博客园博客，原文链接：http://www.cnblogs.com/jiu0821/p/6984851.html，如需转载请自行联系原作者