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 };
