leetcode 41. First Missing Positive

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41. First Missing Positive

 

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• Total Accepted: 69874
• Total Submissions: 288517
• Difficulty: Hard

 

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.

 

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 (M) Missing Number (H) Find the Duplicate Number

 

 

题解转自：

http://blog.csdn.net/sbitswc/article/details/31870975

 

桶排序，每次当A[i]!= i的时候，将A[i]与A[A[i]]交换，直到无法交换位置。终止条件是 A[i]== A[A[i]]。
然后i -> 0 到n走一遍就好了。

 

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