Container with most water
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首先原题描述如下:
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 linei 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.
题目意思就不翻译了,大概是要找到条纵线然后这两条线以及X轴构成的容器能容纳最多的水。
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;
直观的解释是:容积即面积,它受长和高的影响,当长度减小时候,高必须增长才有可能提升面积,所以我们从长度最长时开始递减,然后寻找更高的线来更新候补;
代码如下:
- class Solution {
- public:
- int maxArea(vector<int> &h) {
- // Start typing your C/C++ solution below
- // DO NOT write int main() function
- int res=0;
- int n = h.size();
- int l=0,r=n-1;
- while(l<r)
- {
- res=max(res,min(h[l],h[r])*(r-l));
- if (h[l]<h[r])
- {
- int k=l;
- while(k<r&&h[k]<=h[l])
- k++;
- l=k;
- }
- else
- {
- int k=r;
- while(k>l&&h[k]<=h[r])
- k--;
- r=k;
- }
- }
- return res;
- }
- };
大数据 88 ms A过。
public int maxArea(int[] height) {if (height == null || height.length < 2) {return 0;} int max = 0;int left = 0;int right = height.length - 1; while (left < right) {max = Math.max(max, (right - left) * Math.min(height[left], height[right]));if (height[left] < height[right])left++;elseright--;} return max;}
转自:http://blog.csdn.net/a83610312/article/details/8548519
http://www.programcreek.com/2014/03/leetcode-container-with-most-water-java/
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