42. Trapping Rain Water
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!
Solution:
class Solution {
public:
int trap(vector<int>& height) {
if (height.size() == 0)
return 0;
int res = 0;
size_t l = 0, r = height.size() - 1;
while(l < r) {
int minh = min(height[l], height[r]);
while(l < r && height[l] <= minh)
res += minh, ++ l;
while(l < r && height[r] <= minh)
res += minh, -- r;
}
res += height[l];
return res - accumulate(height.begin(), height.end(), 0);
}
};
本文介绍了一个经典的算法问题——雨水收集。通过给定的地形高度数组,利用双指针技巧计算地形能够承载多少雨水。该算法首先找到左右两边的最高点,并依据较低一侧的高度来累积雨水量。
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