leetcode 11 Container with Most Water
leetcode 11 Container with Most Water
1.题目描述
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 and n is at least 2.2.中文解释：
给定n个非负整数a1,a2,…,an，其中每个代表一个点坐标（i,ai）。
n个垂直线段例如线段的两个端点在（i,ai）和（i,0）。
找到两个线段，与x轴形成一个容器，使其包含最多的水。
备注：你不必倾倒容器。
3.超时的c++算法
当然，谁都可以想到的解法就是暴力匹配，当遇到等差数列的时候当然就超时了！！！
// leetcode11.cpp : 定义控制台应用程序的入口点。
//
#include "stdafx.h"
#include<vector>
using namespace std;
int maxArea(vector<int>& height)
{
int max = 0;
if (height.size() == 0) return 0;
int length = height.size();
int area = 0;
for(int i=0;i<length;i++)
{
for (int j = i; j < length; j++)
{
int high = height[j] > height[i]?height[i]:height[j];
area = high*(j - i);
if (max<area)
{
max = area;
}
}
}
return max;
}
int main()
{
vector<int> array(10);
array.push_back(1);
array.push_back(1);
int area = maxArea(array);
return 0;
}4.正确答案
算法证明：
Here is the proof. Proved by contradiction:
Suppose the returned result is not the optimal solution. Then there must exist an optimal solution, say a container with a_ol and a_or (left and right respectively), such that it has a greater volume than the one we got. Since our algorithm stops only if the two pointers meet. So, we must have visited one of them but not the other. WLOG, let’s say we visited a_ol but not a_or. When a pointer stops at a_ol, it won’t move until
The other pointer also points to a_ol. In this case, iteration ends. But the other pointer must have visited a_or on its way from right end to a_ol. Contradiction to our assumption that we didn’t visit a_or.
The other pointer arrives at a value, say a_rr, that is greater than a_ol before it reaches a_or. In this case, we does move a_ol. But notice that the volume of a_ol and a_rr is already greater than a_ol and a_or (as it is wider and heigher), which means that a_ol and a_or is not the optimal solution – Contradiction!
Both cases arrive at a contradiction.
参考答案：
///C++
int maxArea(vector<int>& height) {
int water = 0;
int i = 0, j = height.size() - 1;
while (i < j) {
int h = min(height[i], height[j]);
water = max(water, (j - i) * h);
while (height[i] <= h && i < j) i++;
while (height[j] <= h && i < j) j--;
}
return water;
}
///C语言参考答案
A bit shorter and perhaps faster because I can use raw int pointers, but a bit longer because I don't have min and max.
int maxArea(int* heights, int n) {
int water = 0, *i = heights, *j = i + n - 1;
while (i < j) {
int h = *i < *j ? *i : *j;
int w = (j - i) * h;
if (w > water) water = w;
while (*i <= h && i < j) i++;
while (*j <= h && i < j) j--;
}
return water;
}python参考答案
class Solution:
def maxArea(self, height):
i, j = 0, len(height) - 1
water = 0
while i < j:
water = max(water, (j - i) * min(height[i], height[j]))
if height[i] < height[j]:
i += 1
else:
j -= 1
return water我的完整工程：
// leetcode11.cpp : 定义控制台应用程序的入口点。
//
#include "stdafx.h"
#include<vector>
using namespace std;
///超时了
int maxArea(vector<int>& height)
{
int max = 0;
if (height.size() == 0) return 0;
int length = height.size();
int area = 0;
for(int i=0;i<length;i++)
{
for (int j = i; j < length; j++)
{
int high = height[j] > height[i]?height[i]:height[j];
area = high*(j - i);
if (max<area)
{
max = area;
}
}
}
return max;
}
///accept
int maxArea2(vector<int>& height)
{
int max = 0;
if (height.size() == 0) return 0;
int length = height.size();
int area = 0;
int l = 0;
int r = length - 1;
while (l < r)
{
int high = height[l] > height[r] ? height[r] : height[l];
max = max > (high * (r - l)) ? max : (high * (r - l));
if (height[l] < height[r])
l++;
else
r--;
}
return max;
}
int main()
{
vector<int> array(0);
array.push_back(1);
array.push_back(1);
int area = maxArea2(array);
return 0;
}