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Wednesday, November 13, 2013

[LeetCode] Candy, Solution

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?
[Thoughts]
蛮好玩的题。感觉用dp简单点。定义Candy[i]为第i个孩子需要给的最少的糖数,
那么
Candy[i] =            Candy[i-1]+1  if ratings[i] > ratings[i-1] 递增序列,后面小孩需要的糖果是前一个小孩的糖果数+1
                           1                   if ratings[i] == ratings[i-1] 直线,按照题意,如果两个小孩rating一样多,后面的小孩可以只拿一个糖
                           Candy[i-1] –1 if ratings[i] < ratings[i-1] 递减序列。这个递推式显然是有缺陷,因为如果递减序列比较长的话,Candy[i]就可能出现负值了,负值显然是没有意义的。比如下图为例:
蓝线是rating变化曲线,数字是Candy[i]的值。基于上面递推式的解(第一行)明显是不合理的。而第二行经过调整的(红色数字),才是最优解。简单的说,就是当遇到一个波谷的时候,调整一下左边的下降序列就好了,但是要注意区分条件,上一个波峰只在有些条件下才需要更改(例一和例二的区别)。
image


[Code]
1:         int candy(vector<int> &ratings) {  
2:            vector<int> candy(ratings.size());  
3:            candy[0] = 1;  
4:            int i =1;  
5:            for (; i < ratings.size(); ++i)  
6:            {  
7:                 if (ratings[i] > ratings[i-1]) //递增  
8:                 {  
9:                      candy[i] = candy[i - 1] + 1;  
10:                 }  
11:                 if (ratings[i] == ratings[i-1]) //平行  
12:                 {  
13:                      candy[i] = 1;  
14:                 }  
15:                 if (ratings[i] < ratings[i - 1]) //递减  
16:                 {  
17:                      candy[i] = candy[i - 1] - 1;  
18:                 }  
19:                 if (i<ratings.size()-1 && ratings[i] < ratings[i-1] && ratings[i] <=ratings[i+1])  
20:                      ReAdjustCandy(ratings, candy, i);  
21:            }  
22:            if (ratings[i-1] < ratings[i-2])  
23:                 ReAdjustCandy(ratings, candy, ratings.size() - 1);  
24:            int total = 0;  
25:            std::for_each(candy.begin(), candy.end(), [&](int n){  
26:                 total += n;  
27:            });  
28:            return total;  
29:       }  
30:       void ReAdjustCandy(vector<int>& ratings, vector<int>& candy, int startIndex)  
31:       {  
32:            int k = startIndex;  
33:            int diff = 1 - candy[k];  
34:            while (k > 0 && ratings[k - 1] > ratings[k])  
35:            {  
36:                 candy[k] = candy[k] + diff;  
37:                 k--;  
38:            }  
39:            if (diff > 0) candy[k] += diff;  
40:       }  



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