1. 题目
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
每次以第一个点为起点,找到后续第一个比起点大于等于的点作为终点。下一次的起点就是上一次的终点。
如果找终点时直到找到末尾也没找到,就将终点设置为当前查找段的最大的点。
确定好段之后,段的首尾是段内的最大和次大点,则直接计算首尾的最大容量,再减去内部的填充点占用即可。
3. 代码
耗时:12ms
class Solution {
public:
// 划分为一段段的处理,每一段是从起点开始,终点是第一个大于等于起点的点。
// 如果终点小于起点,则回退到段内的非起点最高点,作为一段。
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之前的最大点下标
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;
}
};
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