LeetCode:Trapping Rain Water

標籤：algorithm   leetcode   

題目描述：

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.

思路：從左至右依次掃描各列，找出當前列左邊的最大高度hli和右邊的最大高度hri。如果hli>A[i]並且hri>A[i]，則當前列所能盛水的體積為min(hli,hri)- A[i]。

代碼：

int Solution::trap(int A[],int n){    int i,j;    int count = 0;    int * left_height_array = (int *)malloc(sizeof(int)*n);    int left_height_max = 0;    for(i = 0;i < n;i++)    {        left_height_array[i] = left_height_max;        if(A[i] > left_height_max)            left_height_max = A[i];    }    int * right_height_array = (int*)malloc(sizeof(int)*n);    int right_height_max = 0;    for(i = n-1;i >=0;i--)    {        right_height_array[i] = right_height_max;        if(A[i] > right_height_max)            right_height_max = A[i];    }    for(i = 1;i < n-1;i++)    {        int left_height = left_height_array[i];        int right_height = right_height_array[i];        if(left_height > A[i] && right_height > A[i])        {            if(left_height > right_height)                count = count + (right_height - A[i]);            else                count = count + (left_height - A[i]);        }    }    return count;}

LeetCode:Trapping Rain Water