分糖果

LRH1993 · LRH1993 · commit 709083148885 · 2017-10-16T09:43:54.000+08:00
diff --git a/SUMMARY.md b/SUMMARY.md
@@ -212,6 +212,7 @@
   - [Greedy](/algorithm/LeetCode/Greedy.md)
     - [Jump Game](/algorithm/LeetCode/Greedy/Jump-Game.md)
     - [Gas Station](/algorithm/LeetCode/Greedy/Gas-Station.md)
+    - [Candy](/algorithm/LeetCode/Greedy/Candy.md)
 
 ## 设计模式
 
diff --git a/algorithm/LeetCode/Greedy/Candy.md b/algorithm/LeetCode/Greedy/Candy.md
@@ -0,0 +1,51 @@
+## 一、题目
+
+> 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?
+
+有 *N* 个小孩站成一列。每个小孩有一个评级。
+
+按照以下要求，给小孩分糖果：
+
+- 每个小孩至少得到一颗糖果。
+- 评级越高的小孩可以比他相邻的两个小孩得到更多的糖果。
+
+需最少准备多少糖果？
+
+## 二、解题思路
+
+首先我们会给每个小朋友一颗糖果，然后从左到右，假设第i个小孩的等级比第i - 1个小孩高，那么第i的小孩的糖果数量就是第i - 1个小孩糖果数量在加一。再我们从右到左，如果第i个小孩的等级大于第i + 1个小孩的，同时第i个小孩此时的糖果数量小于第i + 1的小孩，那么第i个小孩的糖果数量就是第i + 1个小孩的糖果数量加一。
+
+## 三、解题代码
+
+```java
+public class Solution {
+    public int candy(int[] ratings) {
+        if(ratings == null || ratings.length == 0) {
+            return 0;
+        }
+
+        int[] count = new int[ratings.length];
+        Arrays.fill(count, 1);
+        int sum = 0;
+        for(int i = 1; i < ratings.length; i++) {
+            if(ratings[i] > ratings[i - 1]) {
+                count[i] = count[i - 1] + 1;
+            }
+        }
+
+        for(int i = ratings.length - 1; i >= 1; i--) {
+            sum += count[i];
+            if(ratings[i - 1] > ratings[i] && count[i - 1] <= count[i]) {  // second round has two conditions
+                count[i-1] = count[i] + 1;
+            }
+        }
+        sum += count[0];
+        return sum;
+    }
+}
+```
+
