. - 力扣（LeetCode）

42. 接雨水

困难

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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。

示例 1：

输入：height = [0,1,0,2,1,0,1,3,2,1,2,1]
输出：6
解释：上面是由数组 [0,1,0,2,1,0,1,3,2,1,2,1] 表示的高度图，在这种情况下，可以接 6 个单位的雨水（蓝色部分表示雨水）。 

示例 2：

输入：height = [4,2,0,3,2,5]
输出：9

提示：

• n == height.length
• 1 <= n <= 2 * 104
• 0 <= height[i] <= 105

class Solution {
public:int trap(vector<int>& height) {int n = height.size();vector<int> left(n, 0);vector<int> right(n, 0);int left_max = 0;int right_max = 0;for (int i = 0; i < n; i++) {left_max = std::max(left_max, height[i]);left[i] = left_max;}for (int i = n - 1; i >= 0; i--) {right_max = std::max(right_max, height[i]);right[i] = right_max;}int ret = 0;for (int i = 0; i < n; i++) {ret += (std::min(left[i], right[i]) - height[i]);}return ret;}
};

class Solution {
public:int trap(vector<int>& height) {int n = height.size();int left = 0;int right = n - 1;int area = 0;int max_left = 0;int max_right = 0;while (left < right) {if (height[left] < height[right]) {if (height[left] > max_left) {max_left = height[left];} else {area += (max_left - height[left]);}left++;continue;}if (height[right] > max_right) {max_right = height[right];} else {area += (max_right - height[right]);}right--;}return area;}
};