LeetCode题解–135. Candy
2017-05-04 16:23
288 查看
链接
LeetCode题目:https://leetcode.com/problems/candy/难度:Hard
题目
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?
题目大意是N个小孩站成一排,每个小孩都被分配了一个分数。现在要给他们发糖,需要满足两个条件:一,每个小孩至少分到一块糖;二,分数高的小孩必须比他相邻的小孩分的糖多。求问最少需要发多少块糖?
分析
这题一开始想的是用贪心算法做,先给第一个小孩发1块糖。后面每加一个小孩站在右边,如果他的分数比左边的小孩高,给他比左边小孩发多1块糖;如果分数相等,只给他发1块糖;如果分数低,则是我们需要重点关注的情况,先给他发1块糖,然后向左回溯,检查每个小孩相比其右边小孩是否分数更高但糖不更多,如果是的话,给他发多1块糖。这个思路没问题,但因为时间复杂度是O(n^2),不能通过最后一个测试用例。重新思考了一下,其实不需要回溯。如果遇到的小孩连续都是分数比左边低的,这个分数子序列会变成一个递减序列,我们可以用max_pos记录这个递减序列最左边的位置,max_candy记录这个位置的小孩当前拿了几颗糖。这样一来,每次遇到第i个分数低的小孩,可以把[max_pos+1, i-1]这个区间的小孩发多1块糖,然后再判断max_pos位置的小孩是否需要多得1颗糖。
代码
class Solution { public: int candy(vector<int> &ratings) { int n = (int) ratings.size(); if (n == 0) return 0; vector<int> candies((unsigned long) n, 0); candies[0] = 1; int sum = 1; int max_pos = 0, max_candy = 1; for (int i = 1; i < n; i++) { if (ratings[i] > ratings[i - 1]) { candies[i] = candies[i - 1] + 1; max_pos = i; max_candy = candies[i]; } else if (ratings[i] == ratings[i - 1]) { candies[i] = 1; max_pos = i; max_candy = 1; } else { candies[i] = 1; sum += i - max_pos - 1; if (max_candy < i - max_pos + 1) { max_candy = i - max_pos + 1; sum += 1; } } sum += candies[i]; } return sum; } };
相关文章推荐
- 【LeetCode】135. Candy
- LeetCode 135. Candy
- LeetCode 135. Candy (O(n)时间 O(n)和O(1)空间的两种实现)
- [LeetCode] 135. Candy java
- leetcode题解-135. Candy
- leetcode-135. Candy
- leetcode 135. Candy
- LeetCode 135. Candy
- LeetCode 135. Candy
- [LeetCode]135. Candy
- [leetcode] 【数组】 135. Candy
- [Leetcode] 135. Candy 解题报告
- [Leetcode] 135. Candy
- [leetcode] 135. Candy
- [leetcode] 135. Candy 解题报告
- leetcode-135. Candy
- [算法分析与设计] leetcode 每周一题: 135. Candy
- 135. Candy Leetcode Python
- (Java) LeetCode 135. Candy —— 分发糖果
- [Leetcode] 135. Candy