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LeetCode: Candy

时间:2014-08-23 15:09:00      阅读:157      评论:0      收藏:0      [点我收藏+]

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LeetCode: 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?

地址:https://oj.leetcode.com/problems/candy/

算法:这道题用动态规划可以解决,之前在王道的练习赛上有遇到过,分析见我之前的一篇文章:http://www.cnblogs.com/boostable/p/online_judge_1470_1549_1493_1550.html

代码:

 1 class Solution {
 2 public:
 3     int candy(vector<int> &ratings) {
 4         int n = ratings.size();
 5         vector<int> num_candys(n);
 6         if(n == 0)
 7             return 0;
 8         return subSolution(0,n-1,num_candys,ratings); 
 9     }
10     int subSolution(int s, int e, vector<int> &num_candys,vector<int> &ratings){
11         if(s == e){
12             num_candys[s] = 1;
13             return 1;
14         }else{
15             int total = 0;
16             int mid = (s + e) >> 1;
17             int left = subSolution(s,mid,num_candys,ratings);
18             int right = subSolution(mid+1,e,num_candys,ratings);
19             if(ratings[mid] == ratings[mid+1]){
20                 return left + right;
21             }else if(ratings[mid] > ratings[mid+1]){
22                 if(num_candys[mid] > num_candys[mid+1])
23                     return left + right;
24                 else{
25                     total += (num_candys[mid+1] + 1 - num_candys[mid]);
26                     num_candys[mid] = num_candys[mid+1] + 1;
27                     int totalAdd = 0;
28                     for(int i = mid-1; i >= s; --i){
29                         if(ratings[i] <= ratings[i+1] || (ratings[i] > ratings[i+1] && num_candys[i] > num_candys[i+1]))
30                             break;
31                         else{
32                             total += (num_candys[i+1] + 1 - num_candys[i]);
33                             num_candys[i] = num_candys[i+1] + 1;
34                         }
35                     }
36                     return left + right + total;
37                 }
38             }else{
39                 if(num_candys[mid] < num_candys[mid+1])
40                     return left + right;
41                 else{
42                     total += (num_candys[mid] + 1 - num_candys[mid+1]);
43                     num_candys[mid+1] = num_candys[mid] + 1;
44                     for(int i = mid+2; i <= e; ++i){
45                         if(ratings[i] <= ratings[i-1] || (ratings[i] > ratings[i-1] && num_candys[i] > num_candys[i-1]))
46                             break;
47                         else{
48                             total += (num_candys[i-1] + 1 - num_candys[i]);
49                             num_candys[i] = num_candys[i-1] + 1;
50                         }
51                     }
52                     return left + right + total;
53                 }
54             }
55         }
56     }
57 };

 

LeetCode: Candy

标签:style   blog   http   color   os   io   strong   for   ar   

原文地址:http://www.cnblogs.com/boostable/p/leetcode_candy.html

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