LeetCode 41:First Missing Positive

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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.

 //类似于桶排序，交换数组元素，使得数组中第i位存放数值(i+1);
 //最后遍历数组，寻找第一个不符合此要求的元素，返回其下标。整个过程需要遍历两次数组，复杂度为O(n)。
 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[nums[i]-1], nums[i]);
		 }
		 for (int i= 0; i < n; ++i)
		 {
			 if (nums[i] != i + 1)
				 return i + 1;
		 }
		 return n + 1;
	 }
 };

LeetCode 41:First Missing Positive

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原文地址：http://blog.csdn.net/geekmanong/article/details/50454646