LeetCode刷题之135.分发糖果问题
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LeetCode 之135.分发糖果问题
N个数为一排，每个位置至少加1，相邻两个数评分更高的会得到更多的数，计算所有位置数之和最小。
解题思路：
相邻A与B，A在B左边，若A<B，则B位置加1；若A>B，则A位置加1。
class Solution {
public:
int candy(vector<int>& ratings) {
• vector<int> left(ratings.size(),1);
• vector<int> right(ratings.size(),1);
• //左法则
• for(int i=1;i<ratings.size();i++)
• {
• if(ratings[i]>ratings[i-1]) left[i]=left[i-1]+1;
• }
• int count=0;
• //右法则
• for(int i=ratings.size()-2;i>=0;i--)
• {
• if(ratings[i]>ratings[i+1]) right[i]=right[i+1]+1;
• count+=max(left[i],right[i]);
• }
• count+=max(left[ratings.size()-1],right[ratings.size()-1]);
• return count;
}
};
若将一排数围成一圈，依然遵循上述规则
class Solution {
public:
int candy(vector<int>& ratings) {
• if(ratings.size()==0|| ratings==NULL) return 0;
if(ratings.size()==1) return 1;
vector<int> left(ratings.size(),1);
• vector<int> right(ratings.size(),1);
• //左法则
• for(int i=0;i<ratings.size();i++)
• {
• if(i==0)
{
if(ratings[i]>ratings[ratings.size()-1]) //将ratings[rating.size()-1]放在放 //首位左边。
{left[i]=left[ratings.size()-1]+1;}
}else
{if(ratings[i]>ratings[i-1]) left[i]=left[i-1]+1;}
• }
• int count=0;
• //右法则
• for(int i=ratings.size()-1;i>=0;i--)
• {
• if(i==ratings.size()-1)
{
if(ratings[i]>ratings[0]) //将ratings[0]放在末尾右边
{right[i]=right[0]+1;}
}else
{if(ratings[i]>ratings[i+1]) right[i]=right[i+1]+1;}
• count+=max(left[i],right[i]);
}
• return count;
}
};
TRANSLATE with
标签：count,ratings,right,int,135,left,刷题,LeetCode,size 来源： https://www.cnblogs.com/haipengwu/p/16103026.html