题目
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
2022年12月解法
class Solution { public: int trap(vector& height) { const int c = height.size(); vector leftMax©, rightMax©; for (int i = 1; i < c; i++) { leftMax[i] = max(leftMax[i - 1], height[i - 1]); } for (int i = c - 2; i >= 0; i–) { rightMax[i] = max(rightMax[i + 1], height[i + 1]); } int iSum = 0; for (int i = 0; i < c; i++) { int iNum = min(leftMax[i], rightMax[i]) - height[i]; if (iNum > 0) { iSum += iNum; } } return iSum; } };
2023年8月
class Solution { public: int trap(vector& height) { m_c = height.size(); vector vWaterHeight(m_c); vector<pair<int, int>> vHeightIndex; for (int i = 0; i < m_c; i++) { vHeightIndex.emplace_back(height[i], i); } sort(vHeightIndex.begin(), vHeightIndex.end()); int left, right; left = right = vHeightIndex.back().second; vWaterHeight[left] = vHeightIndex.back().first; for (int i = vHeightIndex.size() - 2; i >= 0; i–) { const auto& [h, inx] = vHeightIndex[i]; if (inx < left) { for (int j = inx; j < left; j++) { vWaterHeight[j] = h; } left = inx; } if (inx > right) { for (int j = inx; j > right; j–) { vWaterHeight[j] = h; } right = inx; } } int iRet = std::accumulate(vWaterHeight.begin(), vWaterHeight.end(), 0) - std::accumulate(height.begin(), height.end(), 0); return iRet; } int m_c; };
其它
视频课程
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https://edu.csdn.net/course/detail/38771
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https://edu.csdn.net/lecturer/6176
测试环境
win7 VS2019 C++17
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