feat: trapping rain water

gp

Author: junbinding

Diff for: stack.trapping-rain-water.py

@@ -0,0 +1,88 @@
+from typing import List
+
+class Solution:
+    """
+    42. 接雨水
+    https://leetcode-cn.com/problems/trapping-rain-water/
+    给定 n 个非负整数表示每个宽度为 1 的柱子的高度图，计算按此排列的柱子，下雨之后能接多少雨水。
+    """
+    # 双指针
+    def trap(self, height: List[int]) -> int:
+        res = 0
+        n = len(height)
+        if n == 0:
+            return res
+
+        l_max = height[0]
+        r_max = height[n - 1]
+        # 定义左指针
+        l = 0
+        # 定义右指针
+        r = n - 1
+
+        while l <= r:
+            # 获取左边最大值
+            l_max = max(l_max, height[l])
+            # 获取右边最大值
+            r_max = max(r_max, height[r])
+
+            # 当左边最大值小于右边最大值，则最多盛水 l_max - height[i]
+            if l_max < r_max:
+                res += l_max - height[l]
+                l += 1
+            else:
+                res += r_max - height[r]
+                r -= 1
+
+        return res
+
+    # 备忘录方式：计算并缓存住每个位置左右两侧的最大高度
+    def trapByNote(self, height: List[int]) -> int:
+        res = 0
+        if not height:
+            return res
+        length = len(height)
+        l_max = [0] * length
+        r_max = [0] * length
+        l_max[0] = height[0]
+        r_max[length - 1] = height[length - 1]
+
+        # 从左向右计算 l_max
+        for i in range(1, length):
+            l_max[i] = max(height[i], l_max[i-1])
+
+        # 从右向左计算 r_max
+        for i in range(length-2, 0, -1):
+            r_max[i] = max(height[i], r_max[i+1])
+
+        # 计算每个高度
+        for i in range(1, length - 1):
+            res += min(l_max[i], r_max[i]) - height[i]
+
+        return res
+
+    # 暴力破解：获取每根柱子可以盛放的水
+    def trapByForce(self, height: List[int]) -> int:
+        length = len(height)
+        res = 0
+        for i in range(1, length - 1):
+            l_max = 0
+            r_max = 0
+            # 获取右侧的最大高度
+            for j in range(i, length):
+                r_max = max(r_max, height[j])
+
+            # 获取左侧的最大高度
+            for k in range(0, i):
+                l_max = max(l_max, height[k])
+
+            # 相对较小的高度减去当前柱子的高度，就是当前柱子可盛放的高度
+            tmp_res = min(l_max, r_max) - height[i]
+            res += tmp_res if tmp_res >= 0 else 0
+
+        return res
+
+
+so = Solution()
+# result is 9
+print(so.trap([0,1,0,2,1,0,1,3,2,1,2,1]))