LeetCode（Array)1431. Kids With the Greatest Number of Candies

﹏ヽ暗。殇╰゛Y 2023-09-23 22:45 59阅读 0赞

# 1.问题 #

There are n kids with candies. You are given an integer array candies, where each candies\[i\] represents the number of candies the ith kid has, and an integer extraCandies, denoting the number of extra candies that you have.

Return a boolean array result of length n, where result\[i\] is true if, after giving the ith kid all the extraCandies, they will have the greatest number of candies among all the kids, or false otherwise.

Note 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: If you give all extraCandies to:
> 
>  *  Kid 1, they will have 2 + 3 = 5 candies, which is the greatest among the kids.
>  *  Kid 2, they will have 3 + 3 = 6 candies, which is the greatest among the kids.
>  *  Kid 3, they will have 5 + 3 = 8 candies, which is the greatest among the kids.
>  *  Kid 4, they will have 1 + 3 = 4 candies, which is not the greatest among the kids.
>  *  Kid 5, they will have 3 + 3 = 6 candies, which is the greatest 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.  
> Kid 1 will always have the greatest number of candies, even if a different kid is given the extra candy.

Example 3:

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

Constraints:

*  n == candies.length
 *  2 <= n <= 100
 *  1 <= candies\[i\] <= 100
 *  1 <= extraCandies <= 50

# 2. 解题思路 #

## 方法1： ##

> 1.定义list  
> 2.定义最大值max的初始值为0  
> 3.candies中的最大值  
> 4.candies数组元素加上extraCandies后比较，如果大于true，小于false  
> 5.返回list值

# 3. 代码 #

## 代码1： ##

class Solution {
        
        public List<Boolean> kidsWithCandies(int[] candies, int extraCandies) {
        
           List<Boolean> list = new ArrayList<>();//1.定义list
            int max =0;//2.定义最大值max的初始值为0
            for (int i = 0; i < candies.length; i++) {
        //3.candies中的最大值
                max = Math.max(candies[i], max);
            }
            for (int i = 0; i < candies.length; i++) {
        
                list.add(candies[i]+extraCandies>=max);//4.candies数组元素加上extraCandies后比较，如果大于true，小于false
            }
            return list;//返回list值  
        }
    }

## 代码2： ##

> 和代码1的解题思路基本相同

public List<Boolean> kidsWithCandies(int[] candies, int extraCandies) {
        
    	List<Boolean> list = new ArrayList<>();
    	int max = 0;
    	for (int i : candies){
        //for each遍历，求最大值
    		if (i > max){
        
    			max = i;
    		}
    	}
    	for (int i = 0; i < candies.length; i++) {
        
    		if (candies[i] + extraCandies >= max) list.add(true);
    		else list.add(false);
    	}
    	return list;
    }