Trapping Rain Water

最新推荐文章于 2024-09-17 16:23:14 发布

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本文介绍了一个经典的算法问题——雨水收集问题。通过一个直观的例子展示如何计算一个由非负整数表示的高度图中能够收集多少雨水。算法采用双指针策略，从两端向中间遍历，利用左右两边的最大高度差来计算能收集的雨水量。

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!

代码如下：

public class TrappingRainWater {

	public int trap(int[] height){
		int ans=0;
		int left=0,right=height.length-1;
		int maxleft=0,maxright=0;
		
		while(left <= right){
			if(height[left] <= height[right]){//类似于短板效应，从短的一侧开始计算
				if(height[left]>=maxleft)	//对比值一直取最大值
					maxleft = height[left];
				else
					ans += maxleft-height[left];
				left++;
			}else{
				if(height[right]>=maxright)	//对比值一直取最大值
					maxright = height[right];
				else
					ans += maxright-height[right];
				right--;
			}
		}
		return ans;
	}
}