Leetcode 135 Candy
2016-11-16 19:31
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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?
分糖果,每个人至少一个,得分较左右高的人要比左右的人分的多。
先是想了DP,后来发现不需要,每个人初始分一个,然后找到单调上升和单调下降的区间,区间转折点取最大值。
程序实现的时候是正一遍,反一遍来的,实现起来比较方便。
class Solution {
public:
int candy(vector<int>& ratings) {
vector<int> dp(ratings.size(), 1);
int res = 0;
for(int i = 1; i < ratings.size();i++) if(ratings[i]>ratings[i-1]) dp[i] = dp[i-1] + 1;
for(int i = ratings.size()-2;i >= 0 ;i--)
{
if(ratings[i] > ratings[i+1]) dp[i] = max(dp[i], dp[i+1] + 1) ;
res += dp[i];
}
return res + dp[ratings.size()-1];
}
};
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?
分糖果,每个人至少一个,得分较左右高的人要比左右的人分的多。
先是想了DP,后来发现不需要,每个人初始分一个,然后找到单调上升和单调下降的区间,区间转折点取最大值。
程序实现的时候是正一遍,反一遍来的,实现起来比较方便。
class Solution {
public:
int candy(vector<int>& ratings) {
vector<int> dp(ratings.size(), 1);
int res = 0;
for(int i = 1; i < ratings.size();i++) if(ratings[i]>ratings[i-1]) dp[i] = dp[i-1] + 1;
for(int i = ratings.size()-2;i >= 0 ;i--)
{
if(ratings[i] > ratings[i+1]) dp[i] = max(dp[i], dp[i+1] + 1) ;
res += dp[i];
}
return res + dp[ratings.size()-1];
}
};
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