LeetCode 268. Missing Number

原题链接在这里：https://leetcode.com/problems/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?

题解：

再刷时，感觉好极了！

与First Missing Positive相似。第一遍把nums[i]放到对应位置上，第二遍看谁不在对应位置上. 若是都在位置上说明missing number 是n, 也就是nums.length.

Time Complexity: O(n). Space: O(1).

AC Java:

public class Solution {
    public int missingNumber(int[] nums) {
        if(nums == null || nums.length == 0){
            return 0;
        }
        for(int i = 0; i<nums.length; i++){
            if(nums[i] >= 0 && nums[i] < nums.length && nums[i] != nums[nums[i]]){
                swap(nums, i, nums[i]);
                i--;
            }
        }
        for(int i = 0; i<nums.length; i++){
            if(nums[i] != i){
                return i;
            }
        }
        return nums.length;
    }
    private void swap(int [] nums, int i, int j){
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
    }
}

AC C++:

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        int len = nums.size();
        for(int i = 0; i < len; i++){
            if(nums[i] >= 0 && nums[i] < len && nums[i] != nums[nums[i]]){
                swap(nums[i], nums[nums[i]]);
                i--;
            }
        }

        for(int i = 0; i < len; i++){
            if(i != nums[i]){
                return i;
            }
        }

        return len;
    }
};

 

第一种方法是 求和，然后挨个减掉，剩余的值就是结果，因为如果是全的，那么sum = n*(n+1)/2.

Time Complexity: O(nums.length). Space: O(1).

第二种方法是从前往后做bit Manipulation, res初始为0，每次保留结果 和 (i+1)^nums[i] 取结果，方法非常巧妙。

Time Complexity: O(nums.length). Space: O(1).

AC Java:

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

跟上Find All Numbers Disappeared in an Array.