1. 題目

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.

For example,
Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped. Thanks Marcos for contributing this image!

2. 思路

每次以第一個點為起點，找到后續第一個比起點大于等于的點作為終點。下一次的起點就是上一次的終點。
如果找終點時直到找到末尾也沒找到，就將終點設置為當前查找段的最大的點。

確定好段之后，段的首尾是段內的最大和次大點，則直接計算首尾的最大容量，再減去內部的填充點占用即可。

3. 代碼

耗時：12ms

class Solution { public:// 劃分為一段段的處理，每一段是從起點開始，終點是第一個大于等于起點的點。// 如果終點小于起點，則回退到段內的非起點最高點，作為一段。int trap(vector<int>& height) {if (height.size() < 3) {return 0; }int sum = 0;int start = 0;int ls = height[start];int end = start + 1;int max_end = start + 1; // start之后, end之前的最大點下標int le = 0;while (end < height.size()) {int cle = height[end];if (cle >= ls) {sum += trap(height, start, end);start = end;ls = cle;end = start + 1;le = 0;max_end = end;continue;} else if (cle > le) {le = cle;max_end = end;}++end;if (end == height.size()) {end = max_end;sum += trap(height, start, end);start = end;end = start + 1;le = 0;max_end = end;}}return sum;}int trap(vector<int>& height, int start, int end) {//cout << "s=" << start << " e=" << end << endl;if (end - start < 2) {return 0;}int sum = (end - start - 1) * min(height[start], height[end]);for (int i = start + 1; i < end; i++) {sum -= height[i];}return sum;} };

總結

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