Trapping Rain Water

标签： C++ 算法 LeetCode 数组

每日算法——leetcode系列

问题 Trapping Rain Water

Difficulty: Hard

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.

<small>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!</small>

class Solution {
public:
    int trap(vector<int>& height) {
        
    }
};

翻译

困住的雨水

难度系数：困难

给定n个代表地图中的高度的非负数的整数，每条的宽度为1，求下雨后能困住多少雨水。
例如：
给定[0,1,0,2,1,0,1,3,2,1,2,1], 返回 6.

思路

• 先找到最高的条柱
• 再两边往中间扫
  1. 扫左边时，如果当前柱左边的最大值是当前柱，就记录
  2. 不是当前柱，刚把水累加

代码

class Solution {
public:
    int trap(vector<int>& height) {
        int maxBarIndex = 0;
        int index = 0;
        for (auto item : height){
            if (item > height.at(maxBarIndex)){
                maxBarIndex = index;
            }
            index++;
        }
        int water = 0;
        for (int i = 0, leftMax = 0; i < maxBarIndex; ++i){
            if (height[i] > leftMax){
                leftMax = height[i];
            } else {
                water += leftMax - height[i];
            }
        }
        int n = static_cast<int>(height.size());
        for (int i = n - 1, rightMax = 0; i > maxBarIndex; --i){
            if (height[i] > rightMax) {
                rightMax = height[i];
            } else {
                water += rightMax - height[i];
            }
        }
        return water;
    }
};

此题还有一种有栈的方式，以后再码。