LeetCode 268 Missing Number (位运算)

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?

Credits:

Special thanks to
@jianchao.li.fighter for adding this problem and creating all test cases.

题目链接：https://leetcode.com/problems/missing-number/

题目分析：这个题目本身不难，最简单办法就是求和减

public class Solution {
public int missingNumber(int[] nums) {
int sum = 0, n = nums.length;
for (int i = 0; i < n; i ++) {
sum += nums[i];
}
return n * (n + 1) / 2 - sum;
}
}
再仔细想一下，其实可以转化成n个数字，其中一个出现了一次，其余的都出现了两次这样一个问题，异或经典问题
public class Solution {
public int missingNumber(int[] nums) {
int ans = nums.length;
for (int i = 0; i < nums.length; i ++) {
ans ^= nums[i];
ans ^= i;
}
return ans;
}
}