题目要求:
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:把出现的数值放到与下标一致的位置,再判断什么位置最先出现不连续的数值。
代码:
class Solution {
public:
int firstMissingPositive(int A[], int n)
{
if(A == NULL || n == 0)
return 1;
for(int i = 0; i < n; ++i)
{
if(A[i] < n && A[i] > 0)
{
if(A[i] != A[A[i] - 1])
{
swap(A[i], A[A[i] - 1]);
--i;
}
}
}
for(size_t i = 0; i < n; ++i)
{
if((A[i] - 1) != i)
return (i + 1);
}
return (n + 1);
}
};