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.
思路:2个指针“遍历比较”所有线段的配对得到面积大小。对于当前区间[left,right],不失一般性,假设height[left]<height[right],那么left++就意味着舍弃了[left,left+1]、...、[left,right-1]这些线段对组成的面积,这是显然的,对于[left,j](j<right),围成的面积=min(height[left],height[j])*(j-left) < height[left]*(right-left),因此这些线段对围成的面积都是可以不用计算的,相当于遍历过,搜索空间进一步减少为[left+1,right]。最后收敛于left==right。
代码:
1 public class Solution { 2 public int maxArea(int[] height) { 3 int maxarea = 0, left = 0, right = height.length - 1; 4 while(left < right) { 5 maxarea = Math.max(maxarea, Math.min(height[left], height[right]) * (right - left)); 6 if (height[left] < height[right]) { 7 left++; 8 } else { 9 right--; 10 } 11 } 12 return maxarea; 13 } 14 }
G
iven 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.