LeetCode 42. Trapping Rain Water--算法题--c++解法

LeetCode 42. Trapping Rain Water–c++解法

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LeetCode题解专栏：LeetCode题解
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大部分题目C++，Python，Java的解法都有。

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题目地址：

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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.

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!

Example:

Input: [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6

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这道题目跟第84题很类似，没做的可以先看这篇文章：LeetCode hard 84. Largest Rectangle in Histogram–python,java 15行,c++ 15行 解法

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这道题目是非常经典的一道题目，我记得字节跳动和百度都出过这道编程题。
最容易想到的是求两边的最大值，然后比较，时间复杂度为O(n)。
C++解法如下：

class Solution {
public:
    int trap(vector<int> &height) {
        if (height.empty()){
            return 0;
        }
           
        int ans = 0;
        int size = height.size();
        vector<int> left_max(size), right_max(size);
        left_max[0] = height[0];
        for (int i = 1; i < size; i++) {
            left_max[i] = max(height[i], left_max[i - 1]);
        }
        right_max[size - 1] = height[size - 1];
        for (int i = size - 2; i >= 0; i--) {
            right_max[i] = max(height[i], right_max[i + 1]);
        }
        for (int i = 1; i < size - 1; i++) {
            ans += min(left_max[i], right_max[i]) - height[i];
        }
        return ans;
    }
};

c++两边往中间夹的解法：

class Solution {
    public int trap(int[] height) {
        // time : O(n)
        // space : O(1)
        if (height.length==0) return 0; 
        int left = 0, right = height.length-1; 
        int leftMax=0, rightMax=0; 
        int ans = 0; 
        while (left < right) {
            if (height[left] > leftMax) leftMax = height[left]; 
            if (height[right] > rightMax) rightMax = height[right];
            if (leftMax < rightMax) {
                ans += Math.max(0, leftMax-height[left]); 
                left++; 
            } else {
                ans += Math.max(0, rightMax-height[right]); 
                right--; 
            }
        }
        return ans; 
    }
}