LeetCode:Trapping Rain Water

浏览数：23 / 时间：2015年06月11日

题目描述：

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：从左到右依次扫描各列，找出当前列左边的最大高度hli和右边的最大高度hri。如果hli>A[i]并且hri>A[i]，则当前列所能盛水的体积为min(hli,hri)- A[i]。

代码：

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;
}

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LeetCode:Trapping Rain Water