[LeetCode] 1431. Kids With the Greatest Number of Candies

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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] 

Constraints:

  • 2 <= candies.length <= 100
  • 1 <= candies[i] <= 100
  • 1 <= extraCandies <= 50

拥有最多糖果的孩子。儿童节来一道儿童题吧。题意是给一个数组表示几个孩子每个人手里目前有几个糖果,同时有一个extraCandies变量。如果这个extraCandies给了任何一个儿童,问是否能让当前这个儿童成为拥有糖果最多的孩子。

时间O(n)

空间O(n)

Java实现

 1 class Solution {
 2     public List<Boolean> kidsWithCandies(int[] candies, int extraCandies) {
 3         int max = 0;
 4         for (int candy : candies) {
 5             max = Math.max(max, candy);
 6         }
 7         List<Boolean> res = new ArrayList<>();
 8         for (int candy : candies) {
 9             if (candy + extraCandies >= max) {
10                 res.add(true);
11             } else {
12                 res.add(false);
13             }
14         }
15         return res;
16     }
17 }

 

javascript实现

 1 /**
 2  * @param {number[]} candies
 3  * @param {number} extraCandies
 4  * @return {boolean[]}
 5  */
 6 var kidsWithCandies = function(candies, extraCandies) {
 7     var max = 0;
 8     var res = [];
 9     for (var i = 0; i < candies.length; i++) {
10         max = Math.max(max, candies[i]);
11     }
12     console.log(max);
13     for (var i = 0; i < candies.length; i++) {
14         if (candies[i] + extraCandies >= max) {
15             res.push(true);
16         } else {
17             res.push(false);
18         }
19     }
20     return res;
21 };

 

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