Re: [閒聊] 每日LeetCode已回收

看板Marginalman作者Rushia (みけねこ的鼻屎)時間2年前 (2023/04/18 00:18)推噓0(0推 0噓 0→)

留言0則, 0人參與討論串296/719 (看更多)

1431. Kids With the Greatest Number of Candies 給你一個陣列 candies ，candies [i] 表示第 i 個小孩有的糖果數量，再給你一個整數 extraCandies 表示額外糖果，返回一個 Boolean 列表，他的值為：若額外糖果全給第 i 個小孩第 i 個小孩的糖果數量會是所有人之中最多的。 Example: 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. 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. 思路： 1.遍歷第一次找出所有小孩中最大的糖果數量。 2.遍歷第二次找出加上額外糖果後，是否會比小孩中最大的糖果數量還多，並設定相應的 Boolean 值。 Java Code： ------------------------------------------------------------------------ class Solution { public List<Boolean> kidsWithCandies(int[] candies, int extraCandies) { int max = candies[0]; for (int candy : candies) { max = Math.max(max, candy); } List<Boolean> res = new ArrayList<>(); for (int candy : candies) { res.add(candy + extraCandies >= max); } return res; } } -------------------------------------------------------------------------- https://i.imgur.com/acHi4CL.png