算法练习（6） —— 贪心算法

习题

本次题目取自leetcode中的 Greedy 栏目中的第135题：
Candy

---

题目很简短：

Description

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?

Expected Input AND Output

Input: A ratings vector
Output: A minimum candies number

思路与代码

• 第一步理解题意， 恐怖如斯， 小孩之间分出了阶级， 阶级越高能拿到的糖果数就越多，阶级最低的只能温饱(至少每个小孩有一颗糖)。而且注意小孩是排成一条线站着的， 不是排成一个圈。给你一个相当于是priority的vector， 让你求能满足条件所给的最少的糖(吝啬)。
• 满足第一条。先把所有的孩子分配一颗糖。然后可以分成2步，第一步从头到尾搜，第二步从尾到头搜。每次只比较一个方向。优先级比旁边那个高，就等于它的糖果数+1
• 其实这题挺简单的，贪心算法讲究的都只是算法正确就能解决，所以演算算法的正确性才是最难的一部分吧

---

代码如下：
“`
#include
using namespace std;

class Solution {
public:
int candy(vector& ratings) {
if (ratings.size() <= 1)
return 1;

    int minCandies = 0;    int* candies = new int[ratings.size()];    for (int i = 0; i < ratings.size(); i++)        candies[i] = 1;    for (int i = 1; i < ratings.size(); i++)        if (ratings[i] > ratings[i - 1])            candies[i] = candies[i - 1] + 1;    for (int i = ratings.size() - 1; i > 0; i--)        if (ratings[i - 1] > ratings[i])            candies[i - 1] = (candies[i] + 1 > candies[i - 1]) ? candies[i] + 1 : candies[i - 1];    for (int i = 0; i < ratings.size(); i++)        minCandies += candies[i];    return minCandies;}

};
“`