Leetcode–Trapping Rain Water

The Problem:

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.

Notes:

  1. This tricky problem I tried many ways but got stuck. You just need a right perspective to see it.
  2. The area of water can be viewed as combination of all water grids, and each water stripe at index k can be calculated as the min of its highest left wall and highest right wall, multiplies grid width 1, then minus the brick height A[k].

Java Solution:

public class Solution {
    public int trap(int[] A) {
        if(A.length<=1) return 0;
        int[] leftHigh = new int[A.length];
        int[] rightHigh = new int[A.length];
        int pre = 0;
        for(int i = 0; i < A.length; i++){
            if(A[i] >pre){
                leftHigh[i] = A[i];
                pre = A[i];
            }
            else{
                leftHigh[i] = pre;
            }
        }
        pre = 0;
        for(int i = A.length-1; i>=0; i--){
              if(A[i] >pre){
                rightHigh[i] = A[i];
                pre = A[i];
            }
            else{
                rightHigh[i] = pre;
            }
        }
        int area = 0;
        for(int i = 0; i < A.length; i++){
            area+= Math.min(leftHigh[i], rightHigh[i]) - A[i];
        }
        return area;
    }
}
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