leetcode42：接雨水

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。

输入: [0,1,0,2,1,0,1,3,2,1,2,1]
输出: 6

 动态编程-->O(n) O(n)

public int trap(int[] height) {
    if (height == null || height.length == 0) 
        return 0;
    int ans = 0;
    int size = height.length;
    int[] left_max = new int[size];
    int[] right_max = new int[size];
    left_max[0] = height[0];
    for (int i = 1; i < size; i++) {
        left_max[i] = Math.max(height[i], left_max[i - 1]);
    }
    right_max[size - 1] = height[size - 1];
    for (int i = size - 2; i >= 0; i--) {
        right_max[i] = Math.max(height[i], right_max[i + 1]);
    }
    for (int i = 1; i < size - 1; i++) {
        ans += Math.min(left_max[i], right_max[i]) - height[i];
    }
    return ans;
}

单调栈-->O(n) O(n)

class Solution:
    def trap(self, height: List[int]) -> int:
        stack = []
        stack.append(0)
        res = 0
        for i in range(1, len(height)):
            while stack and height[stack[-1]] <= height[i]:
                x = stack.pop()
                if not stack:
                    break
                res += (min(height[stack[-1]], height[i]) - height[x]) * (i - stack[-1] - 1)
            stack.append(i) 
        return res

双指针

public int trap(int[] height) {
    int left = 0, right = height.length - 1;
    int ans = 0;
    int left_max = 0, right_max = 0;
    while (left < right) {
        if (height[left] < height[right]) {
            if (height[left] >= left_max) {
                 left_max = height[left];
            } else {
                 ans += (left_max - height[left]);
            }
            ++left;
        } else {
            if (height[right] >= right_max) {
                right_max = height[right];
            } else {
                ans += (right_max - height[right]);
            }
             --right;
        }
    }
    return ans;
}