Create 135._Candy.md

Author: 0xkookoo

docs/Leetcode_Solutions/Python/135._Candy.md

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+# 135. Candy
+
+**<font color=red>难度: Hard</font>**
+
+## 刷题内容
+
+> 原题连接
+
+* https://leetcode.com/problems/candy/
+
+> 内容描述
+
+```
+
+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?
+
+Example 1:
+
+Input: [1,0,2]
+Output: 5
+Explanation: You can allocate to the first, second and third child with 2, 1, 2 candies respectively.
+Example 2:
+
+Input: [1,2,2]
+Output: 4
+Explanation: You can allocate to the first, second and third child with 1, 2, 1 candies respectively.
+             The third child gets 1 candy because it satisfies the above two conditions.
+```
+
+## 解题方案
+
+> 思路 1
+******- 时间复杂度: O(N)******- 空间复杂度: O(N)******
+
+
+
+- 首先我们从左往右边看，如果当前child的rating比它左边child的rating大，那么当前child分的candy肯定要比它左边多1
+- 然后从右往左边看，如果当前child的rating比它右边child的rating大，那么当前child分的candy肯定要比它右边多1
+
+我们知道这两种情况我们都必须满足，所以最后我们取两者中的大者（这样才能同时满足两种情况）,beats 24.83%
+
+```python
+class Solution:
+    def candy(self, ratings):
+        """
+        :type ratings: List[int]
+        :rtype: int
+        """
+        left2right = [1] * len(ratings)
+        right2left = [1] * len(ratings)
+        for i in range(1, len(ratings)):
+            if ratings[i] > ratings[i-1]:
+                left2right[i] = left2right[i-1] + 1
+        for i in range(len(ratings)-2, -1, -1):
+            if ratings[i] > ratings[i+1]:
+                right2left[i] = right2left[i+1] + 1
+        res = 0
+        for i in range(len(ratings)):
+            res += max(left2right[i], right2left[i])
+        return res
+```
+
+> 思路 2
+******- 时间复杂度: O(N)******- 空间复杂度: O(N)******
+
+刚才用了2个array，其实我们可以只用一个array， beats 28.77%
+
+```python
+class Solution:
+    def candy(self, ratings):
+        """
+        :type ratings: List[int]
+        :rtype: int
+        """
+        res = [1] * len(ratings) # also compatable with [] input
+        # left scan
+        for i in range(1, len(ratings)):
+            if ratings[i] > ratings[i-1]:
+                res[i] = res[i-1] + 1
+        # right scan
+        for i in range(len(ratings)-2, -1, -1):
+            if ratings[i] > ratings[i+1]:
+                res[i] = max(res[i+1]+1, res[i])
+        return sum(res)
+```
+
+
+> 思路 3
+******- 时间复杂度: O(N)******- 空间复杂度: O(1)******
+
+其实我们还有另外一种思路，可以将空间降到O(1)
+
+假设：
+```
+input: [2, 4, 6, 8, 7, 6, 5, 4, 3, 2, 7, 8, 9, 8, 7, 6, 5, 4, 3]
+index: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10,11,12,13,14,15,16,17,18,19]
+
+```
+
+我们先直接给第一个孩子1颗candy，然后继续往下遍历，我们发现第2个孩子的rating比第一个孩子的多，于是我们给第二个孩子2颗candy，
+以此类推，第3个孩子3颗candy，第4个孩子4颗candy，这时候我们发现第5个孩子的rating比第4个孩子的rating要小，
+所以第5个孩子的candy肯定要比第4个孩子的少才行，但是不能直接给第5个孩子1颗candy这么简单，因为说不定后面孩子的rating比第5个孩子更小呢？
+如果第5个孩子只有1颗candy，后面的孩子岂不是没有candy了？这不符合题目要求，所以我们暂时继续走下去，我们发现一共降序走了6步，此时来到了第11个孩子，
+rating终于又变成升序了，这时候我们知道了，第5个孩子最少需要给6颗candy了，然后第6，7，8，9，10个孩子依次给5，4，3，2，1颗candy，总数就是6*(1+6)/2
+但是问题来了，第5个孩子给了6颗candy，第4个孩子应该要比第5个孩子的candy更多呀， 于是我们补给第4个孩子3颗candy，此时第4个孩子有7颗candy
+（比第5个孩子的6颗多，符合题目要求）。继续走下去，第11个孩子给2颗candy，第12个孩子给3颗candy，第13个孩子给4颗candy，接下来又是降序了，
+又是降序走了6步，一样的原理，第14,15,16,17,18,19个孩子分别给6,5,4,3,2,1颗candy，发现第13个孩子需要补3颗candy（现在7颗candy）
+
+于是我们的candy的总和为
+
+```
+input: [2, 4, 6, 8, 7, 6, 5, 4, 3, 2, 7, 8, 9, 8, 7, 6, 5, 4, 3]
+candy: [1, 2, 3, 7, 6, 5, 4, 3, 2, 1, 2, 3, 7, 6, 5, 4, 3, 2, 1]
+index: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10,11,12,13,14,15,16,17,18,19]
+
+一共67颗candy
+```
+
+编码实现, beats 59.40%
+
+```python
+class Solution:
+    def candy(self, ratings):
+        """
+        :type ratings: List[int]
+        :rtype: int
+        """
+        if not ratings:
+            return 0
+        res, prev, desc_cnt = 1, 1, 0
+        for i in range(1, len(ratings)):
+            if ratings[i] >= ratings[i-1]:
+                if desc_cnt > 0:
+                    res += desc_cnt * (desc_cnt + 1) / 2 # arithmetic progression
+                    if desc_cnt >= prev:
+                        res += desc_cnt - prev + 1
+                    desc_cnt = 0
+                    prev = 1
+                # 当前child的rating和它左边child的rating一样，直接给1颗candy就满足了，否则需要给prev+1颗candy
+                prev = 1 if ratings[i] == ratings[i-1] else prev + 1 
+                res += prev
+            else:
+                desc_cnt += 1
+        if desc_cnt > 0: # if we were descending at the end
+            res += desc_cnt * (desc_cnt + 1) / 2 # arithmetic progression
+            if desc_cnt >= prev:
+                res += desc_cnt - prev + 1
+        return int(res)
+```
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