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:
What is the minimum candies you must give?
/**
* Abstract each slope, and sort by the length of the slop. For those slop whose length is one, then add one from
* it bottom. The candy is the sum from the bottom to ceil.
* When abstract the slope, we should node the length of the length of the before slop.
* 1. if the ratings equals to their neighbors? I think it should be in just one.
*
* */
//9:38->the equals should get just 1,not equal than its neighbor.10:00 finished.
class Solution {
public:
int candy(vector<int> &ratings) {
int minCandy = 0,tmp=0;
int i=0,len = ratings.size();
// those who's rating equals to it's neighbors should be just one
vector<int> candies(len,1);
int candy = 1;
candies[0] = 1;
// from left to right
for(i=1;i<len;i++)
{
if(ratings[i]>ratings[i-1])
{
candies[i] = candies[i-1]+1;
}
}
// from right to left
for(i=len-2;i>=0;i--)
{
if(ratings[i]>ratings[i+1])
{
tmp = candies[i+1]+1;
if(tmp>candies[i])
{
candies[i] = tmp;
}
}
}
// add the sum
for(i=0;i<len;i++)
{
minCandy += candies[i];
}
return minCandy;
}
};原文地址：http://blog.csdn.net/nan327347465/article/details/39156857
