42. Trapping Rain Water *HARD*

標籤：

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. 

int trap(int a[], int n) {    int result = 0;    //find the highest value/position    int maxHigh = 0;    int maxIdx = 0;    for(int i=0; i<n; i++){        if (a[i] > maxHigh){            maxHigh = a[i];            maxIdx = i;        }    }    //from the left to the highest postion    int prevHigh = 0;    for(int i=0; i<maxIdx; i++){        if(a[i] > prevHigh){            prevHigh = a[i];        }        result += (prevHigh - a[i]);    }    //from the right to the highest postion    prevHigh=0;    for(int i=n-1; i>maxIdx; i--){        if(a[i] > prevHigh){            prevHigh = a[i];        }        result += (prevHigh - a[i]);    }    return result;}

* The idea is:
* 1) find the highest bar.
* 2) traverse the bar from left the highest bar.
* becasue we have the highest bar in right, so, any bar higher than its right bar(s) can contain the water.
* 3) traverse the bar from right the highest bar.
* becasue we have the highest bar in left, so, any bar higher than its left bar(s) can contain the water.

42. Trapping Rain Water *HARD*