即判斷A[i]是否等于i
**41. 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.
代碼如下：
class Solution {
public:
int firstMissingPositive(vector<int>& nums) {
int n = nums.size();
if(n==0)
return 1;
for(int i=0;i<n;i++)
{
if(nums[i]>0&&nums[i]<n) //要把1,2,3… 放在最前面云挟，這樣有利于后續(xù)的判斷
{
if(nums[i]-1==i)
continue;
else if(nums[nums[i]-1]!=nums[i])
{
int temp = nums[nums[i]-1];
nums[nums[i]-1] = nums[i];
nums[i] = temp;
i--;
}
}
}
for(int i=0;i<n;i++)
{
if(nums[i]!=i+1)
return i+1;
}
return n+1;
}
};
