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描述

Given n non-negative integers representing an elevation map where the width of each bar is 1, computehow 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.

分析：

对于每个柱子，找到其左右两边最高的柱子，该柱子能容纳的面积就是 min(max_left,max_right) - height。所以，
1. 从左往右扫描一遍，对于每个柱子，求取左边最大值；
2. 从右往左扫描一遍，对于每个柱子，求最大右值；
3. 再扫描一遍，把每个柱子的面积并累加。

code：

/*trapping train water v2*/
    int trap(vector<int> &height)
    {
        int result=0;
        int n = height.size();
        int *left_flag = new int[n];
        int *right_flag = new  int[n];
        memset(left_flag, INT_MAX, n*sizeof(int));
        memset(right_flag, INT_MAX, n* sizeof(int));
        left_flag[0] = 0;
        right_flag[n - 1] = 0;

        //扫描左边
        for (int i = 1; i < n; i++)
        {
            if (height[i - 1]>left_flag[i - 1])
            {
                left_flag[i] = height[i - 1];
            }
            else
            {
                left_flag[i] = left_flag[i - 1];
            }
        }
        //扫描右边
        for (int i = n - 2; i >= 0; i--)
        {
            if (height[i + 1] > right_flag[i + 1])
                right_flag[i] = height[i + 1];
            else
                right_flag[i] = right_flag[i + 1];
        }

        for (size_t i = 0; i < n; i++)
        {
            int compare = min(left_flag[i], right_flag[i]) - height[i];
            result += (compare>0 ? compare : 0);
        }

        delete[] left_flag;
        delete[] right_flag;
        return result;
    }

相关题目: (M) Container With Most Water

                 Product of Array Except Self

[leetcode]trapping rain water

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原文地址：http://www.cnblogs.com/panweishadow/p/4781427.html