LeetCode 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. 

Solution 1:

分析某个柱子，发现该柱子上水的高度和其左右两边的最高柱子有关，设该柱子左边所有柱子中最高的为leftmax，右边所有柱子中最高的为rightmax，如果min(leftmax, rightmax) 大于该柱子的高度，那么该柱子上可以蓄水为min(leftmax, rightmax) - 该柱子高度，如果min(leftmax, rightmax) <= 该柱子高度，则该柱子上没有蓄水。

可以从后往前扫描一遍数组求得某个柱子右边的最高柱子，从前往后扫描一遍数组求得某个柱子左边的最高柱子, 然后按照上面的分析可以求得所有的蓄水量。

public int trap(int[] A) {
    int n = A.length, waterSum = 0;
    int[] rightMax = new int[n];
    int maxH = 0;
    for(int i=n-1; i>=0; i--) {
        rightMax[i] = maxH;
        maxH = Math.max(maxH, A[i]);
    }
    maxH = 0;
    for(int i=0; i<n; i++) {
        int minH = Math.min(maxH, rightMax[i]);
        if(A[i] < minH) {
            waterSum += minH - A[i];
        }
        maxH = Math.max(maxH, A[i]);
    }
    return waterSum;
}

 

Reference:

http://www.cnblogs.com/TenosDoIt/p/3812880.html