Leetcode: Trapping Rain Water

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!

方法1：模拟

array [0,1,0,2,1,0,1,3,2,1,2,1]的矩阵如上图所示，是一个3*12的矩阵，然后从矩阵的第一行开始扫描，遇到两个1就加上这两个1之间的水量，只要扫描一遍矩阵就得到水量了，代码如下，小集合通过，大集合memory不够用。

Judge Small: Accepted!

Judge Large: Memory Limit Exceeded

int trap(int A[], int n) {        // Note: The Solution object is instantiated only once.        if(A==NULL || n<1)return 0;    int highest = A[0];for(int i=1;i<n;i++)if(A[i]>highest)highest=A[i];int ** matrix = new int*[highest];for(int i=0;i<highest;i++){matrix[i]=new int[n];memset(matrix[i],0,sizeof(int)*n);}for(int i = 0; i < n; i++)for(int j = 1; j <= A[i]; j++)matrix[highest-j][i]=1;int water = 0;int left = -1;for(int i = 0; i < highest; i++){left = -1;for(int j = 0; j < n; j++){if(matrix[i][j]==1){if(left==-1)left = j;else{water += j-left-1;left = j;}}}}for(int i=0;i<highest;i++)delete[] matrix[i];delete[] matrix;return water;    }

既然memory不够，就把int矩阵改成bool吧，这样空间就由int的4byte缩小到bool的1 byte，但是还是Memory Limit Exceeded

int trap(int A[], int n) {        // Note: The Solution object is instantiated only once.        if(A==NULL || n<1)return 0;    int highest = A[0];for(int i=1;i<n;i++)if(A[i]>highest)highest=A[i];bool ** matrix = new bool*[highest];for(int i=0;i<highest;i++){matrix[i]=new bool[n];memset(matrix[i],0,sizeof(bool)*n);}for(int i = 0; i < n; i++)for(int j = 1; j <= A[i]; j++)matrix[highest-j][i]=1;int water = 0;int left = -1;for(int i = 0; i < highest; i++){left = -1;for(int j = 0; j < n; j++){if(matrix[i][j]){if(left==-1)left = j;else{water += j-left-1;left = j;}}}}for(int i=0;i<highest;i++)delete[] matrix[i];delete[] matrix;return water;    }

但仔细想想感觉这道题应该是扫一遍就能得到结果的。。。

对某个值A[i]来说，能trapped的最多的water取决于在i之前最高的值leftMostHeight[i]和在i右边的最高的值rightMostHeight[i]（均不包含自身）。

如果min(left,right) > A[i]，那么在i这个位置上能trapped的water就是min(left,right) – A[i]。

有了这个想法就好办了，第一遍从左到右计算数组leftMostHeight，第二遍从右到左计算rightMostHeight。

时间复杂度是O(n)。

int trap(int A[], int n) {        // Note: The Solution object is instantiated only once.        if(A==NULL || n<1)return 0;    int maxheight = 0;vector<int> leftMostHeight(n);for(int i =0; i<n;i++){leftMostHeight[i]=maxheight;maxheight = maxheight > A[i] ? maxheight : A[i];}maxheight = 0;vector<int> rightMostHeight(n);for(int i =n-1;i>=0;i--){rightMostHeight[i] = maxheight;maxheight = maxheight > A[i] ? maxheight : A[i];}int water = 0;for(int i =0; i < n; i++){int high = min(leftMostHeight[i],rightMostHeight[i])-A[i];if(high>0)water += high;}return water;    }

精简下代码：

class Solution {public:    int trap(vector<int>& height) {        if(height.size() < 2) return 0;        int sz = height.size();        vector<int> left(sz, 0);        vector<int> right(sz, 0);        for(int i = 1; i < sz; i++){            left[i] = max(left[i-1], height[i-1]);            right[sz-i-1] = max(right[sz-i], height[sz-i]);        }        int water = 0;        for(int i = 1; i < sz; i++){            int minh = min(left[i],right[i]);            water += minh > height[i] ? minh - height[i] : 0;        }        return water;    }};