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LeetCode42 Trapping Rain Water

题目:

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.

技术分享

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.  (Hard)

 分析:

首先考虑对每个位置向左向右找到最小值,然后与本位置的值比较添加存水结果,可以做出来,时间复杂度O(n^2);

考虑利用空间优化时间,开两个数组,分别存当前位置向左的最小值和向右的最小值。

一次从左往右遍历更新leftHeight数组, 一次从右向左遍历更新rightHeight数组,最后一次遍历算利用上述方法算result,只不过这次可以直接从数组里读左右最小值结果。

代码:

 1 class Solution { 2 public: 3     int trap(vector<int>& height) { 4         int leftMax[height.size()]; 5         leftMax[0] = 0; 6         for (int i = 1; i < height.size(); ++i) { 7             leftMax[i] = max(height[i - 1], leftMax[i - 1]); 8         } 9         int rightMax[height.size()];10         rightMax[height.size() - 1] = 0;11         for (int i = height.size() - 2; i >= 0; --i) {12             rightMax[i] = max(height[i + 1], rightMax[i + 1]);13         }14         int result = 0;15         for (int i = 0; i < height.size(); ++i) {16             if (min(leftMax[i], rightMax[i]) - height[i] > 0) {17                 result += (min(leftMax[i], rightMax[i]) - height[i]);18             }19         }20         return result;21     }22 };

上述方法将时间复杂度优化到了O(n),但利用了额外的空间,空间复杂度也提高到了O(n)。

进一步优化,可以考虑Two pointers的思路。

两根指针分别指向头和尾,并维护两个值,表示从左向右到p1的最大值leftHeight和从右向左到p2的最大值RightHeight。

两个最大值中较小的向中间移动,遇到更大的值更新leftHeight或rightHeight,遇到较小的值更新result。

这样可以做到时间复杂度O(n),空间复杂度O(1)。

代码:

 1 class Solution { 2 public: 3     int trap(vector<int>& height) { 4         if (height.size() < 2) { 5             return 0; 6         } 7         int left = 0, right = height.size() - 1; 8         int leftHeight = height[0], rightHeight = height[height.size() - 1]; 9         int result = 0;10         while (left < right) {11             if (leftHeight <= rightHeight) {12                 left++;13                 if (height[left] < leftHeight) {14                     result += (leftHeight - height[left]);15                 }16                 else {17                     leftHeight = height[left];18                 }19             }20             else {21                 right--;22                 if (height[right] < rightHeight) {23                     result += (rightHeight - height[right]);24                 }25                 else {26                     rightHeight = height[right];27                 }28             }29         }30         return result;31     }32 };

 

 

LeetCode42 Trapping Rain Water