LeetCode 11 Container With Most Water (C,C++,Java,Python)
2015-05-07 22:56
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Problem:
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.
Solution:
下面以例子: [4,6,2,6,7,11,2] 来讲解。1.首先假设我们找到能取最大容积的纵线为 i , j (假定i<j),那么得到的最大容积 C = min( ai , aj ) * ( j- i) ;
2.下面我们看这么一条性质:
①: 在 j 的右端没有一条线会比它高! 假设存在 k |( j<k && ak > aj) ,那么 由 ak> aj,所以 min( ai,aj, ak) =min(ai,aj) ,所以由i, k构成的容器的容积C' = min(ai,aj ) * ( k-i) > C,与C是最值矛盾,所以得证j的后边不会有比它还高的线;
②:同理,在i的左边也不会有比它高的线;
这说明什么呢?如果我们目前得到的候选: 设为 x, y两条线(x< y),那么能够得到比它更大容积的新的两条边必然在 [x,y]区间内并且 ax' > =ax , ay'>= ay;
3.所以我们从两头向中间靠拢,同时更新候选值;在收缩区间的时候优先从 x, y中较小的边开始收缩;
题目大意:
在X轴上给定一些竖线,竖线有长度,求两条竖线与X轴构成的一个容器能容纳最多的水的面积解题思路:
如Solution,时间复杂度O(n)Java源代码(用时373ms):
public class Solution { public int maxArea(int[] height) { int Max=-1,l=0,r=height.length-1; while(l<r){ int area=(height[l]<height[r]?height[l]:height[r])*(r-l); Max=Max<area?area:Max; if(height[l]<height[r]){ int k=l; while(l<r && height[l]<=height[k])l++; }else{ int k=r; while(l<r && height[r]<=height[k])r--; } } return Max; } }
C语言源代码(用时12ms):
int maxArea(int* height, int heightSize) { int Max=-1,area,l=0,r=heightSize-1,k; while(l<r){ area=(height[l]<height[r]?height[l]:height[r])*(r-l); Max=Max<area?area:Max; if(height[l]<height[r]){ k=l; while(l<r && height[l]<=height[k])l++; }else{ k=r; while(l<r && height[r]<=height[k])r--; } } return Max; }
C++源代码(用时30ms):
class Solution { public: int maxArea(vector<int>& height) { int Max=-1,l=0,r=height.size()-1,area,k; while(l<r){ area=(height[l]<height[r]?height[l]:height[r])*(r-l); Max=Max<area?area:Max; if(height[l]<height[r]){ k=l; while(l<r && height[l]<=height[k])l++; }else{ k=r; while(l<r && height[r]<=height[k])r--; } } return Max; } };
Python源代码(用时152ms):
class Solution: # @param {integer[]} height # @return {integer} def maxArea(self, height): Max=-1;l=0;r=len(height)-1 while l<r: area=(height[l] if height[l]<height[r] else height[r])*(r-l) Max=Max if Max>area else area if height[l]<height[r]: k=l while l<r and height[l]<=height[k]:l+=1 else: k=r while l<r and height[r]<=height[k]:r-=1 return Max
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