[leetcode] Trapping Rain Water

最新推荐文章于 2021-11-04 09:56:06 发布

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本文介绍了一种计算降雨后地形中积水总量的算法。通过找出地形中的最高点，并分别从左右两侧进行扫描，累计可积水区域的高度差来计算总的积水体积。提供两种实现方案：一种基于两次遍历的方法，另一种采用双指针简化处理。

From : https://leetcode.com/problems/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!

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(M) Container With Most Water (M) Product of Array Except Self

solution 1:

1.先找到最大处；

2.从左端到最高处（这时右边一定被挡住了），如果左边可以挡住（左边有更高的地方），那么当前位置就可以trap water；

3.从右边到最高处，同上。

class Solution {
public:
    int trap(vector<int>& height) {
        int maxP = 0, len = height.size(), h = 0, water = 0;
        if(!len) return 0;
        for(int i=0; i<len; ++i) {
            if(h < height[i]) {
                maxP = i;
                h = height[i];
            }
        }
        
        h = height[0];
        for(int i=1; i<maxP; ++i) {
            int hi = height[i];
            hi > h ? h = hi : water += h-hi;
        }
        h = height[len-1];
        for(int i=len-2; i>maxP; --i) {
            int hi = height[i];
            hi > h ? h = hi : water += h-hi;
        }
        return water;
    }
};

Solution 2：

对上面方法简化，双指针；从两端开始，右边更高，那么只要左边能被挡住，就能trap water；反之亦然。

class Solution {
public:
    int trap(vector<int>& height) {
        int left = 0, right = height.size()-1, water = 0;
        while(left < right) {
            int lv = height[left], rv = height[right];
            if(lv > rv) {
                for(right = right-1; right > left && height[right] < rv; --right) {
                    water += rv-height[right];
                }
            } else {
                for(left=left+1; left < right && height[left] < lv; ++left) {
                    water += lv-height[left];
                }
            }
        }
        return water;
    }
};