Lintcode_接雨水

原题目地址

描述

给出 n 个非负整数，代表一张X轴上每个区域宽度为 1 的海拔图, 计算这个海拔图最多能接住多少（面积）雨水。

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.

样例

• 样例 1:
  输入: [0,1,0]
  输出: 0

• 样例 2:

  输入: [0,1,0,2,1,0,1,3,2,1,2,1]
  输出: 6

挑战

• O(n) 时间, O(1) 空间
• O(n) 时间, O(n) 空间也可以接受

思路

• 计算柱子I能承接多少水，其实就是寻找左边最大的极值LeftMax[I]，右边最大的极值RightMax[I]；
• 在水不漏掉的情况下，即左右两边的最大极值大于height[I]
• 得到每根柱子能承接的水量 min(LeftMax[I], RightMax[I]) - height[I]

代码

在实际操作中，将一次寻找左边最大值的遍历 和 最后的计算合并

class Solution:
    """
    @param heights: a list of integers
    @return: a integer
    """

    def trapRainWater(self, heights):
        length = len(heights)

        if length < 2:
            return 0

        leftMax = []
        rightMax = []

        # for i in range(length):
        #     if i == 0:
        #         leftMax.append(0)
        #     elif i == 1:
        #         leftMax.append(heights[0])
        #     else:
        #         leftMax.append(max(leftMax[i - 1], heights[i - 1]))

        for i in range(length - 1, -1, -1):
            if i == length - 1:
                rightMax.append(0)
            elif i == length - 2:
                rightMax.append(heights[length - 1])
            else:
                rightMax.append(max(heights[i + 1], rightMax[length - i - 2]))


        # rightMax.reverse()
        # print(rightMax)

        rain = 0
        for i in range(length):
            if i == 0:
                leftMax.append(0)
                rain += 0
                continue
            elif i == 1:
                leftMax.append(heights[0])
            else:
                leftMax.append(max(leftMax[i - 1], heights[i - 1]))

            if leftMax[i] > heights[i] and rightMax[length-i-1] > heights[i]:
                rain += min(leftMax[i], rightMax[length-i-1]) - heights[i]

        # print(leftMax)
        return rain


if __name__ == '__main__':
    print(Solution().trapRainWater([1, 3, 4, 5, 2, 4, 7, 4, 3]))