41. First Missing Positive

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.

难度：hard

题目：
给定一无排序的整数数组，找出丢失的最小正整数。

注意：
算法时间复杂度为O(n)空间复杂度为常量。

思路：
清除所有原数组中不在范围［1,n(数组长度)]的值。并将所有index + 1 = nums[index]的数组归位。归位过程中相等的两个元素不用交换（避免无限循环）

Runtime: 5 ms, faster than 100.00% of Java online submissions for First Missing Positive.

class Solution {
    public int firstMissingPositive(int[] nums) {
        if (null == nums || nums.length < 1) {
            return 1;
        }
        
        int t = 0, n = nums.length;
        for (int i = 0; i < n; i++) {
            // make it as zero to avoid array index overflow
            if (nums[i] <= 0 || nums[i] > n) {
                nums[i] = 0;
            } else if ((i + 1) != nums[i] && nums[nums[i] - 1] != nums[i]) {
                t = nums[nums[i] - 1];
                nums[nums[i] - 1] = nums[i];
                nums[i--] = t;
            }
        }
        
        for (int i = 0; i < nums.length; i++) {
            if ((i + 1) != nums[i]) {
                return i + 1;
            }
        }
        
        return n + 1;
    }
}