# 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?

### 迭代版​

// Candy// 时间复杂度O(n)，空间复杂度O(n)public class Solution {    public int candy(int[] ratings) {        final int n = ratings.length;        final int[] increment = new int[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] = Math.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] = Math.max(inc++, increment[i]);            else                inc = 1;        }        // 初始值为n，因为每个小朋友至少一颗糖        int sum = n;        for (int i : increment) sum += i;        return sum;    }};

### 递归版​

// Candy// 备忘录法，时间复杂度O(n)，空间复杂度O(n)// java.lang.StackOverflowErrorpublic class Solution {    public int candy(int[] ratings) {        final int[] f = new int[ratings.length];        int sum = 0;        for (int i = 0; i < ratings.length; ++i)            sum += solve(ratings, f, i);        return sum;    }    int solve(int[] ratings, int[] f, int i) {        if (f[i] == 0) {            f[i] = 1;            if (i > 0 && ratings[i] > ratings[i - 1])                f[i] = Math.max(f[i], solve(ratings, f, i - 1) + 1);            if (i < ratings.length - 1 && ratings[i] > ratings[i + 1])                f[i] = Math.max(f[i], solve(ratings, f, i + 1) + 1);        }        return f[i];    }}