Trapping Rain Water

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本文详细介绍了如何通过算法计算给定高度数组所表示的地形在降雨后能存留的水量。包括核心概念、具体步骤和代码实现。通过实例展示如何在数组中找到最高柱子，利用左右最大高度来计算每根柱子能存留的水量。

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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!

Analysis: For each element in the given array, the water it can contain is determined by its leftMaxBarValue and rightMaxBarValue.

if min(leftMaxBarValue, rightMaxBarValue)>currentBarValue, waterContained+=min(leftMaxBarValue, rightMaxBarValue)-currentBarValue

In order to get the left and right bounds for each element, we can scan the array two times:

1. FInd the index and value of the highest bar in the array.

2. Scan the array, the element on the left of the maxBarIndex should have maxBarValue as their rightMaxBarValue. Their leftMaxBarValue can be updated and determined on the way. Same logic works for the right part of the array, except that we should scan from the end to the maxBarIndex.

public class Solution {
    public int trap(int[] A) {
        int maxBarValue=0, maxBarIndex=-1;
        for(int i=0; i<A.length; i++) {
            if(A[i] >= maxBarValue) {   // must be >= here
                maxBarValue = A[i];
                maxBarIndex = i;
            }
        }
        if(maxBarValue==0 || maxBarIndex==-1) return 0;
        
        int[] leftBound = new int[A.length];
        int[] rightBound = new int[A.length];
        int leftMax=0, rightMax=0;
        for(int j=0; j<maxBarIndex; j++) {  // 0 -> maxBarIndex
            leftBound[j] = leftMax;
            rightBound[j] = maxBarValue;
            if(A[j]>leftMax) leftMax=A[j];
        }
        leftBound[maxBarIndex] = -1;
        rightBound[maxBarIndex] = -1;
        for(int k=A.length-1; k>=maxBarIndex+1; k--) {    // A.length-1 -> maxBarIndex
            leftBound[k] = maxBarValue;
            rightBound[k] = rightMax;
            if(A[k]>rightMax) rightMax=A[k];
        }
        
        int water = 0;
        for(int m=0; m<A.length; m++) {
            int lowBarValue = Math.min(leftBound[m], rightBound[m]);
            if(lowBarValue>A[m]) water+=lowBarValue-A[m];
        }
        return water;
    }
}