LeetCode-135. Candy

LeetCode-135. Candy

There are N children standing in a line. Each child is assigned a rating value.

You are giving candies to these children subjected to the following requirements:

Each child must have at least one candy.Children with a higher rating get more candies than their neighbors.

What is the minimum candies you must give?

Example 1:

Input: [1,0,2]Output: 5Explanation: You can allocate to the first, second and third child with 2, 1, 2 candies respectively.

Example 2:

Input: [1,2,2]Output: 4Explanation: You can allocate to the first, second and third child with 1, 2, 1 candies respectively. The third child gets 1 candy because it satisfies the above two conditions.

题解：

贪心算法。

从前往后遍历如果第i + 1个权重大于i，则sum[i + 1] = sum[i] + 1。

从后往前遍历，如果 第i个权重大于i + 1并且sum[i]小于等于sum[i + 1]，那么sum[i] = sum[i + 1] + 1。

class Solution {public: int candy(vector& ratings) { int n = ratings.size(); int res = 0; vector sum(n, 1); for (int i = 0; i < n - 1; i++) { if (ratings[i + 1] > ratings[i]) { sum[i + 1] = sum[i] + 1; } } for (int i = n - 2; i >= 0; i--) { if (ratings[i] > ratings[i + 1] && sum[i] <= sum[i + 1]) { sum[i] = sum[i + 1] + 1; } } for (int i = 0; i < n; i++) { res += sum[i]; } return res; }};

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