题目描述

Given the array candies and the integer extraCandies, where candies[i] represents the number of candies that the ith kid has.

For each kid check if there is a way to distribute extraCandies among the kids such that he or she can have the greatest number of candies among them. Notice that multiple kids can have the greatest number of candies.

Example 1:

Input: candies = [2,3,5,1,3], extraCandies = 3
Output: [true,true,true,false,true] 
Explanation: 
Kid 1 has 2 candies and if he or she receives all extra candies (3) will have 5 candies --- the greatest number of candies among the kids. 
Kid 2 has 3 candies and if he or she receives at least 2 extra candies will have the greatest number of candies among the kids. 
Kid 3 has 5 candies and this is already the greatest number of candies among the kids. 
Kid 4 has 1 candy and even if he or she receives all extra candies will only have 4 candies. 
Kid 5 has 3 candies and if he or she receives at least 2 extra candies will have the greatest number of candies among the kids. 

Example 2:

Input: candies = [4,2,1,1,2], extraCandies = 1
Output: [true,false,false,false,false] 
Explanation: There is only 1 extra candy, therefore only kid 1 will have the greatest number of candies among the kids regardless of who takes the extra candy.

Example 3:

Input: candies = [12,1,12], extraCandies = 10
Output: [true,false,true]

题目思路

• 思路一、数组中存放的是孩子们消灾手中的糖果数，现有额外的糖果，问孩子们现在手中糖果加上一部分额外糖果能否和目前孩子们手中糖果最大数目持平。
  第一步：找到目前孩子们手中最大糖果数
  第二步：循环判断每个孩子手中糖果加上额外糖果是否大于最大糖果数

class Solution {
public:
    vector<bool> kidsWithCandies(vector<int>& candies, int extraCandies) {
        vector<bool> res;
        int mc = candies[0];
        // 找到 kids 目前手中的最大 candies
        for(int i = 1; i < candies.size(); ++i)
        {
            if(candies[i] > mc) mc=candies[i];   
        }
        
        // kids 手中已知 greatest candies 保存在 mc 中
        for(int i = 0; i < candies.size(); ++i)
        {
            if(candies[i]+extraCandies >= mc){
                res.push_back(true);
            }
            else{
                res.push_back(false);
            }
        }
        return res;
    }
};

总结展望