leetcod--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.

解题思路：先求一遍不缺少情况下数组的和，然后减去缺少情况下数组的元素，差值就是缺少的元素。

#include <iostream>
#include<vector>
using namespace std;int missingNumber(vector<int>& nums)
{int n=nums.size();cout<<n<<endl;int sum=(0+n)*(n+1)/2;for(int i=0; i<n; i++)  sum-=nums[i];return sum;
}
int main()
{vector<int> num;num.push_back(0);num.push_back(1);num.push_back(2);num.push_back(4);int s=missingNumber(num);cout<<s<<endl;}

拓展：如果要找到缺失的2个数呢？

package leedcode;import java.math.*;
import java.util.ArrayList;
import java.util.Arrays;public class test {public static ArrayList missingNumber(int[] nums) {ArrayList List = new ArrayList();Arrays.sort(nums);for (int i = 0; i < nums.length; i++) {if (nums[i] != i)List.add(i);}return List;}// 主方法public static void main(String[] args) {int[] s = { 0, 2, 4, 5, 6 };ArrayList r = missingNumber(s);System.out.println(r.get(1));}
}

不太好的一点就是开辟了新的空间。。。。。。。。。。

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