Leetcode Missing Number

标签：

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?

解题思路：

方法一：求和法。算出0..n 的和然后减去数组中值的和，就是Missing Number

方法二：Bit Manipulation，用异或。类似Leetcode Single Number。

既然0到n之间少了一个数，我们将这个少了一个数的数组合0到n之间完整的数组异或一下，那么相同的数字都变为0了，剩下的就是少了的那个数字了。

Java code:

方法一：

public class Solution {
    public int missingNumber(int[] nums) {
       int n = nums.length;
       int sum = n * (n + 1) / 2;
       for(int i = 0; i < n; i++) {
           sum -= nums[i];
       }
       return sum;
    }
}

方法二：

public class Solution {
    public int missingNumber(int[] nums) {
        int result = 0;
        for(int i = 0; i < nums.length; i++){
            result ^= (i+1) ^ nums[i];
        }
        return result;
    }
}

Reference:

1. http://www.cnblogs.com/grandyang/p/4756677.html

2. https://leetcode.com/discuss/61256/java-missing-number-solution-o-1-space-and-n-time-complexity

Leetcode Missing Number

标签：

原文地址：http://www.cnblogs.com/anne-vista/p/4899674.html