268. Missing Number

标签：大小   constant   solution   public   i+1   time   NPU   ant   ssi   

Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.

Example 1:

Input: [3,0,1]
Output: 2

Example 2:

Input: [9,6,4,2,3,5,7,0,1]
Output: 8

Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?

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

注意考虑都在的情况（返回数组长）

class Solution {
    public int missingNumber(int[] nums) {
        int k1 = 0, k2 = 0;
        for(int i = 0; i < nums.length; i++){
            k1 += nums[i];
            k2 += (i+1);
        }
        return k2 - k1;
    }
}

或者计算原本应该有的值大小，再减去数组之和就是miss掉的。

268. Missing Number

标签：大小   constant   solution   public   i+1   time   NPU   ant   ssi   

原文地址：https://www.cnblogs.com/wentiliangkaihua/p/11702723.html