算法精粹(algorithm-essentials)

感谢soulmachine@github提供内容
## Candy


### 描述

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?


### 分析




### 迭代版

{% if book.java %}
{% codesnippet "./code/candy-1."+book.suffix, language=book.suffix %}{% endcodesnippet %}
{% endif %}

{% if book.cpp %}
```cpp
// Candy
// 时间复杂度O(n),空间复杂度O(n)
class Solution {
public:
int candy(vector &ratings) {
const int n = ratings.size();
vector increment(n);

// 左右各扫描一遍
for (int i = 1, inc = 1; i < n; i++) {
if (ratings[i] > ratings[i - 1])
increment[i] = max(inc++, increment[i]);
else
inc = 1;
}

for (int i = n - 2, inc = 1; i >= 0; i--) {
if (ratings[i] > ratings[i + 1])
increment[i] = max(inc++, increment[i]);
else
inc = 1;
}
// 初始值为n,因为每个小朋友至少一颗糖
return accumulate(&increment[0], &increment[0]+n, n);
}
};
```
{% endif %}


### 递归版

{% if book.java %}
{% codesnippet "./code/candy-2."+book.suffix, language=book.suffix %}{% endcodesnippet %}
{% endif %}

{% if book.cpp %}
```cpp
// Candy
// 备忘录法,时间复杂度O(n),空间复杂度O(n)
// @author fancymouse (http://weibo.com/u/1928162822)
class Solution {
public:
int candy(const vector& ratings) {
vector f(ratings.size());
int sum = 0;
for (int i = 0; i < ratings.size(); ++i)
sum += solve(ratings, f, i);
return sum;
}
int solve(const vector& ratings, vector& f, int i) {
if (f[i] == 0) {
f[i] = 1;
if (i > 0 && ratings[i] > ratings[i - 1])
f[i] = max(f[i], solve(ratings, f, i - 1) + 1);
if (i < ratings.size() - 1 && ratings[i] > ratings[i + 1])
f[i] = max(f[i], solve(ratings, f, i + 1) + 1);
}
return f[i];
}
};
```
{% endif %}