Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.
Example 1
Input: [3,0,1]
Output: 2
Example 2
Input: [9,6,4,2,3,5,7,0,1]
Output: 8
Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?
根据高斯求和公式，1+2+3+4+……+ n = n (n+1) /2
先求出总和，再减去数组所有元素的和，就可以得到缺失的那个数了。
class Solution {
public int missingNumber(int[] nums) {
int total = nums.length * (nums.length + 1) / 2;
int sum = 0;
for (int num : nums) {
sum += num;
}
return total - sum;
}
}可以使用异或运算符找出缺失的那个数。异或运算符有如下特性：
- a⊕0=a
- a⊕a=0
- a⊕b⊕a=(a⊕a)⊕b=0⊕b=b
由于数组的索引从 0 到 n-1，数组的元素从 0 到 n 各不相同且只缺失一个。 那么只需要初始化一个数 n，然后异或每一个索引和对应的元素，最后的结果就是缺失的那个数。
class Solution {
public int missingNumber(int[] nums) {
int missing = nums.length;
for (int i = 0; i < nums.length; i++) {
missing ^= i ^ nums[i];
}
return missing;
}
}