[LeetCode41] First Missing Positive

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.

Analysis:

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。

Java

public int firstMissingPositive(int[] A) {
        for(int i=0;i<A.length;i++){
        	while(A[i]!=i+1){
        		if(A[i]<=0 ||A[i]>=A.length|| A[i]==A[A[i]-1]) break;
        		int temp = A[i];
        		A[i] = A[A[i]-1];
        		A[temp-1] = temp;
        	}
        	
        }
        for(int i=0;i<A.length;i++){
        	if(A[i]!=i+1)
        		return i+1;
        }
        return A.length+1;
    }

c++

int firstMissingPositive(int A[], int n) {
        int i=0;
    while(i<n){
        if(A[i]!=i+1 && A[i]>=1 &&A[i]<=n && A[A[i]-1] !=A[i])
            swap(A[i],A[A[i]-1]);
        else
            i++;
    }
    for(int i=0; i<n;i++){
        if(A[i]!=i+1)
            return i+1;
    }
    return n+1;
    }


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