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.
O(n) 时间,并且不能开辟数组。我们通过交换数组中的值,尽量使得A[i]=i+1,这样,当交换完后,再扫描一下数组,如果A[i]!=i+1,则i+1为第一个missing的正整数。
class Solution {
public:
int firstMissingPositive(int A[], int n) {
if(n<=0)return 1;
for(int i=0;i<n;i++){
if(A[i]>0&&A[i]<=n&&A[i]!=i+1&&A[A[i]-1]!=A[i]){
swap(A[i],A[A[i]-1]);
i--;
}
}
for(int i=0;i<n;i++){
if(A[i]!=i+1)return i+1;
}
return n+1;
}
};