Leetcode: Missing Number

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?

Best Solution: Bit manipulation

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.

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

 

Summation approach:

public int missingNumber(int[] nums) { //sum
    int len = nums.length;
    int sum = (0+len)*(len+1)/2;
    for(int i=0; i<len; i++)
        sum-=nums[i];
    return sum;
}

 

 

 

因为输入数组是0,1,2,...，n。 把nums[i]放到i的位置上.  nums[i] != i的即为missing number.

注意6-9行不可先令 temp = nums[i]

public class Solution {
    public int missingNumber(int[] nums) {
        int res = nums.length;
        for (int i=0; i<nums.length; i++) {
            if (nums[i] < nums.length && nums[nums[i]] != nums[i]) {
                int temp = nums[nums[i]];
                nums[nums[i]] = nums[i];
                nums[i] = temp;
                i--;
            }
        }
        for (int i=0; i<nums.length; i++) {
            if (nums[i] != i) res = i;
        }
        return res;
    }
}