[Leetcode 41] First Missing Positive
2017-07-05 21:56
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Question
Given an unsorted integer array, find the first missing positive integer.
思路
给定的数组大小为n;
对数组桶排序后, 中间没有缺少正数的话, 那么 nums[i]==i+1 成立。否则为缺少的数。
桶排序算法的时间复杂度为 O(n), 是满足题意对时间的要求的。
-代码如下:
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.
思路
给定的数组大小为n;
对数组桶排序后, 中间没有缺少正数的话, 那么 nums[i]==i+1 成立。否则为缺少的数。
桶排序算法的时间复杂度为 O(n), 是满足题意对时间的要求的。
-代码如下:
class Solution { public: int firstMissingPositive(vector<int>& nums) { int maxn=1, minn=1; int i=0; while(i<nums.size()){ // 注意交换仅在 nums[i]是正数, 且 nums[i]!= i+1 //nums[nums[i]-1]!=nums[i] 交换目标必须是不合要求的了,这个很重要,不然会陷入死循环 //nums[i]<=nums.size() 大于nums.size()的忽略 if(nums[i]>0 && nums[i]!= i+1 && nums[i]<=nums.size() && nums[nums[i]-1]!=nums[i]){ swap(nums[nums[i]-1],nums[i]); } else ++i; } for(int i=0; i<nums.size(); ++i){ if(nums[i]!=i+1) return i+1; } return nums.size()+1; } };
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