268. Missing Number(Java)

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?

代码1：

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

// 相比于上面的方法可以防止溢出class Solution {    public int missingNumber(int[] nums) {        int sum = nums.length;        for (int i = 0; i < nums.length; i ++)             sum += i - nums[i];          return sum;    }}

代码2：利用A ^ A = 0，A ^ A ^ B = B，且满足结合律和交换律

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

代码3：若数组有序，可以考虑用二分搜索

class Solution {    public int missingNumber(int[] nums) {        Arrays.sort(nums);        int left = 0, right = nums.length, mid = (left + right) / 2;        while (left < right) {            // mid为本来应该的平均数            mid = (left + right) / 2;            // 若实际偏大，则抽走了小的数，则抽走的数在左边            if (nums[mid] > mid)                 right = mid;            // 若实际偏小，则抽走了大的数，则抽走的数在右边            else left = mid + 1;        }        return left;    }}