LeetCode-6:接雨水

Posted 2022-03-11 il_持之以恒_li

原题为：

参考代码为：

class Solution(object):
    def trap(self, height):
        """
        :type height: List[int]
        :rtype: int
        """
        n = len(height)
        dp = [0] * n
        left = 0
        right = n - 1
        max_v = height[0]
        max_index = 0
        for i in range(n):
            if height[i] > max_v:
                max_v = height[i]
                max_index = i

        for i in range(0, max_index + 1):
            if height[i] > height[left]:
                left = i
                dp[i] = height[i]
            else:
                dp[i] = height[left]

        for i in range(n - 1, max_index, -1):
            if height[i] > height[right]:
                right = i
                dp[i] = height[i]
            else:
                dp[i] = height[right]

        sum_1 = 0
        for i in range(n):
            sum_1 += dp[i] - height[i]
        
        return sum_1

个人思路如下：

拿上图举个例子吧！如果我们把上图柱子的高度和(图二)减去原来柱子的高度和(图一)是不是就是所要求的结果了呀！那么怎样才能得到图二的数据呢？小编觉得这样，先遍历原来列表，找到列表中柱子最高的下标索引max_index，之后，再重新遍历列表。
当下标索引小于等于max_index的情况下,
如果(left初始值为0)height[left]<height[i]，那么，left = i，dp[i] = height[i]；
否则，dp[i] = height[left]，(这里dp是一个列表，长度为height的长度，初始值为[0]*height的长度)下标索引大于max_index的情况和上述判断差不多。
执行效率：

改进：

class Solution(object):
    def trap(self, height):
        """
        :type height: List[int]
        :rtype: int
        """
        n = len(height)
        left = 0
        right = n - 1
        max_v = height[0]
        max_index = 0
        for i in range(n):
            if height[i] > max_v:
                max_v = height[i]
                max_index = i

        sum_1 = 0
        for i in range(0,n):
            if i <= max_index:
                if height[i] > height[left]:
                    left = i
                    dp = height[i]
                else:
                    dp = height[left]
                sum_1 += dp - height[i]
            else:
                right_index = n-(i-max_index)
                if height[right_index] > height[right]:
                    right = right_index
                    dp = height[right_index]
                else:
                    dp = height[right]
                sum_1 += dp - height[right_index]

        return sum_1

执行效率：