Given an unsorted integer array, find the smallest missing positive integer.

Example 1:

Input: [1,2,0]
Output: 3

Example 2:

Input: [3,4,-1,1]
Output: 2

Example 3:

Input: [7,8,9,11,12]
Output: 1

Note:

Your algorithm should run in O(n) time and uses constant extra space.

因为不能额外使用空间,所以不能用HashTable,利用原本的数组记录状态。将值为正整数的数字放到对应值-1的下标位置。

注意无限循环:比如[1,1],所以nums[nums[i]-1]==nums[i]的情况要排除。

class Solution {
public int firstMissingPositive(int[] nums) {
int tmp;
int i = 0;
while(i < nums.length){
if(nums[i] < nums.length && nums[i] > 0 && nums[i] != i+1 && nums[nums[i]-1]!=nums[i]){
//put nums[i] at position of nums[i]-1
tmp = nums[i];
nums[i] = nums[tmp-1];
nums[tmp-1] = tmp;
}
else{
i++;
}
} for(i = 0; i < nums.length; i++){
if(nums[i] != i+1) break;
}
return i+1;
}
}

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