leetcod--Missing Number

来源：互联网发布：淘宝店家骂人怎么办编辑：程序博客网时间：2024/06/16 05:54

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。。。。。。。。。。

1 0