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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.
题目大意：求一系列点之间的最大面积
第一种方法：时间复杂度(O(n^2))
1 public int maxArea(int[] height) { 2 int area = 0; 3 int min = 0; 4 for(int i=0; i<height.length; i++){ 5 for(int j=i+1; j<height.length; j++){ 6 if(height[i] == height[j]) 7 area = Math.max(area, height[i]*(j-i)); 8 else{ 9 min = Math.min(height[i],height[j]); 10 area = Math.max(area, min*(j-i)); 11 } 12 } 13 } 14 return area; 15 }
第二种方法：
1 public int maxArea(int[] height) { 2 int area = 0; 3 int begin = 0; 4 int end = height.length-1; 5 while(begin<end){ 6 area = Math.max(area, (end-begin)*Math.min(height[begin],height[end])); 7 if(height[begin]<height[end]) 8 begin++; 9 else 10 end--; 11 } 12 return area; 13 }
[LeetCode]-011-Container_With_Most_Water
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原文地址：http://www.cnblogs.com/lianliang/p/5403602.html
