LeetCode Missing Number



Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.

For example,
Given nums = [0, 1, 3] return 2.

Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?

题意 ：给出一个由0到n之间整数组成的数组，数组个数为n,每个元素不相等，问数组中缺少哪个数，要求时间复杂度为O(n),空间复杂度为O(1)

思路：用置换的方法，在每次循环时，获取当前下标的数组元素，在将tmp赋值给nums[tmp]前，将nums[tmp]放在临时变量中，如果循环操作，下去越界或者与起始位置重合

代码如下

class Solution {
     public int missingNumber(int[] nums)
     {
       for (int i = 0; i < nums.length; i++)
       {
           int tmp = nums[i];
           nums[i] = -1;
           while (tmp != -1 && tmp < nums.length) {
               int t = nums[tmp];
               nums[tmp] = tmp;
               tmp = t;
           }
       }

         int res = -1;
         for (int i = 0; i < nums.length; i++)
         {
             if (nums[i] != i)
             {
                 res = i;
                 break;
             }
         }

         if (res == -1) res = nums.length;

         return res;
     }
}