Trapping Rain Water 收集雨水 C/C++

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. Thanks Marcos for contributing this image!

题目大意：给以一个数组，每个元素表示长方体的高，这些长方体宽度为1，求这个数组的长方体组成的图形能蓄多少水

解题思路：当前位置能蓄水要求两边的高度都高于当前位置，所以我们扫描到当前位置，只需判断当前位置是否对蓄的水有贡献。

当前位置宽度为1最终的贡献值，为左边高度的最大值和右边高度的最大值，取其中最小的，然后和当前高度相减，如果大于0，则这个值就是贡献值

C

int trap(int* height, int heightSize) {
    int level = 0, water = 0;
    while (heightSize--) {
        int lower = *height < height[heightSize] ? *height++ : height[heightSize];
        if (lower > level) level = lower;
        water += level - lower;
    }
    return water;

}

C++：

class Solution {
public:
    int trap(vector<int>& height) {
        int water = 0;
        int size = height.size();
        for (int i = 1; i < size - 1; i++) {
            int max_left = 0, max_right = 0;
            for (int j = i; j >= 0; j--) { //Search the left part for max bar size
                max_left = max(max_left, height[j]);
            }
            for (int j = i; j < size; j++) { //Search the right part for max bar size
                max_right = max(max_right, height[j]);
            }
            water += min(max_left, max_right) - height[i];
        }
        return water;
    }
};