LintCode_362 Trapping Rain Water

最新推荐文章于 2021-01-05 12:22:47 发布

Bernini_buffalo

最新推荐文章于 2021-01-05 12:22:47 发布

阅读量238

点赞数

CC 4.0 BY-SA版权

分类专栏： LintCode

本文链接：https://blog.youkuaiyun.com/qq_31552435/article/details/52113794

LintCode 专栏收录该内容

58 篇文章

订阅专栏

本文介绍了一种计算给定地形图中雨后积水体积的方法。通过寻找地形中的最高点，并分别计算左右两侧能容纳的水量来得出结果。挑战在于实现时间和空间效率高的算法。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

Trapping Rain Water

Example

Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

Challenge

O(n) time and O(1) memory

O(n) time and O(n) memory is also acceptable.

先找出最高点，然后分别计算left的和与right的和，因为在最高点两边，所以互不影响。

class Solution {
public:
    /**
     * @param heights: a vector of integers
     * @return: a integer
     */
    int trapRainWater(vector<int> &heights) {
        // write your code here
        if (heights.size() <= 2)
            return 0;
        heights.push_back(0);
        heights.insert(heights.begin(), 0);
        int max_h = 0;
        int ind;
        int sum_left = 0;
        int bar_left = 0;
        int sum_right = 0;
        int bar_right = 0;
        
        for (int i = 0; i < heights.size(); i++) {
            if (heights[i] > max_h) {
                max_h = heights[i];
                ind = i;
            }
        }
        for (int i = 0; i < ind; i++) {
            if (heights[i] > bar_left) {
                bar_left = heights[i];
            } else {
                sum_left+=(bar_left - heights[i]);
            }
        }
        for (int i = heights.size() - 1; i > ind; i--) {
            if (heights[i] > bar_right) {
                bar_right = heights[i];
            } else {
                sum_right+=(bar_right - heights[i]);
            }
        }
        return sum_left + sum_right;
    }
};