LeetCode 之水箱问题

1. Container with most water

Given n non-negative integers a1, a2, ..., an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.

Note: You may not slant the container.

容积总是由最短板决定，每次移动短板找出最大容积

c++

int maxArea(vector<int> &height) {    int start =0;        int end = height.size()-1;       int maxV = INT_MIN;        while(start<end)        {          int contain = min(height[end], height[start]) * (end-start);          maxV = max(maxV, contain);          if(height[start]<= height[end])          {            start++;          }          else          {            end--;          }        }        return maxV;  }

java

public int maxArea(int[] height) {if(height.length<2) return 0;int maxV = 0;int start =0;int end = height.length-1;while(start<end){int area = Math.min(height[start], height[end])*(end-start);maxV = Math.max(maxV, area);if(height[start]>height[end])end--;else {start++;}}return maxV;    }

2. 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!

思路，坐标的最大左右边相减就是该坐标点容积

1. 从左到右扫描，找出每个坐标的最大左值

2. 从右到左扫描，找出每个坐标最大右值

3. 累加每个容积。

c++

int trap(int A[], int n) {    if(n<2) return 0;    int *maxL = new int[n], *maxR = new int[n];    int maxLR = A[0];    maxL[0] = 0;    //from left to right cal max lvalue    for(int i=1; i<n-1; i++){        maxL[i] = maxLR;        if(A[i]>maxLR){            maxLR = A[i];        }    }    maxLR = A[n-1];    maxR[n-1] = 0;    int ttrap = 0, ctrap = 0;    for(int i=n-2; i>0;i--){        maxR[i] = maxLR;        ctrap = min(maxL[i], maxR[i]) - A[i];        if(ctrap>0)            ttrap+= ctrap;        if(maxLR < A[i])            maxLR = A[i];    }    delete maxL, maxR;    return ttrap;}

java

public int trap(int[] A) {if(A.length<2) return 0;int len = A.length;int []maxL = new int [len];int maxLR = A[0];maxL[0] = 0;for(int i=1;i<len;i++){maxL[i] = maxLR;if(A[i]>maxLR)maxLR = A[i];}maxLR = A[len-1];int []maxR = new int[len];maxR[len-1] = 0;int trap = 0;for(int i=len-2;i>=0;i--){maxR[i] = maxLR;int ctrap = Math.min(maxL[i], maxR[i])-A[i];if(ctrap>0) trap+=ctrap;if(A[i]>maxLR)maxLR = A[i];}return trap;    }