Candy--LeetCode

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?

这道题用到的思路和Trapping
Rain Water是一样的，用动态规划。基本思路就是进行两次扫描，一次从左往右，一次从右往左。第一次扫描的时候维护对于每一个小孩左边所需要最少的糖果数量，存入数组对应元素中，第二次扫描的时候维护右边所需的最少糖果数，并且比较将左边和右边大的糖果数量存入结果数组对应元素中。这样两遍扫描之后就可以得到每一个所需要的最最少糖果量，从而累加得出结果。方法只需要两次扫描，所以时间复杂度是O(2*n)=O(n)。空间上需要一个长度为n的数组，复杂度是O(n)。代码如下：

int candy(vector<int> &ratings) 
{
    vector<int> candy(ratings.size(),1);
    int sum,i;
    for(i=1;i<ratings.size();i++)
    {
       if(ratings[i] > ratings[i-1])
         candy[i] = candy[i-1]+1;
    }
    
    sum = candy[ratings.size()-1];
    for(i=ratings.size()-2;i>=0;i--)
    {
       int cur =1;
       if(ratings[i] > ratings[i+1])
         cur = candy[i+1]+1;
       sum += max(cur,candy[i]);
       candy[i] = cur;
    }
}

如果上面的那个不好理解，看下面的代码

int candy(vector<int> &ratings) 
{
    vector<int> candy(ratings.size(),1);
    int sum,i;
    for(i=1;i<ratings.size();i++)
    {
       if(ratings[i] > ratings[i-1])
         candy[i] = candy[i-1]+1;
    }
  
    sum = candy[ratings.size()-1];
    for(i=ratings.size()-2;i>=0;i--)
    {
       int cur =1;
       if(ratings[i] > ratings[i+1] && candy[i] < candy[i+1])
        	candy[i] = candy[i+1]+1;	
       sum += candy[i];
       
    }
    
    return sum;
}

相当于左右开弓，分别计算最小值，然后就可以了，但是需要注意的是

if(ratings[i] > ratings[i+1] && candy[i] < candy[i+1]) //后半部分的判断尤其重要