268. Missing Number

题目：

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?

链接： http://leetcode.com/problems/missing-number/

题解：

求missing number。首先的想法是先求数列和，再减去数组里的数。这样的话要小心写法，有可能会overflow。我的写法就是会溢出的那种。

Time Complexity - O(n)， Space Complexity - O(1)

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

也可以用Bit Manipulation

Time Complexity - O(n)， Space Complexity - O(1)

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

Reference:

https://leetcode.com/discuss/53802/c-solution-using-bit-manipulation

https://leetcode.com/discuss/56174/3-different-ideas-xor-sum-binary-search-java-code

https://leetcode.com/discuss/53937/simple-c-codes

https://leetcode.com/discuss/53790/1-lines-ruby-python-java-c

https://leetcode.com/discuss/53871/java-simplest-solution-o-1-space-o-n-time

https://leetcode.com/discuss/58647/line-simple-java-bit-manipulate-solution-with-explaination

268. Missing Number

原文：http://www.cnblogs.com/yrbbest/p/5022828.html