268. Missing Number

作者: 番茄晓蛋 | 来源:发表于2016-10-10 13:41 被阅读9次

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    Difficulty: Medium

    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?

    public int missingNumber(int[] nums) {
            /* Ref:
              1.https://leetcode.com/discuss/58647/line-simple-java-bit-manipulate-solution-with-explaination
              2. https://leetcode.com/discuss/56174/3-different-ideas-xor-sum-binary-search-java-code
              
             The basic idea is to use XOR operation. We all know that a^b^b =a, which means two xor operations with the same number will eliminate the number and reveal the original number. 
             
             In this solution, I apply XOR operation to both the index and value of the array. In a complete array with no missing numbers, the index and value should be perfectly corresponding( nums[index] = index), so in a missing array, what left finally is the missing number.
    
            */
            int xor = 0, i = 0;
            for (i=0; i<nums.length; i++) {
                xor = xor ^ i ^ nums[i];
            }
            return xor ^ i;
        }
    

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