【题目】
木桶短板,能装的最多水,由最短的那条决定
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 the line i
is at (i, ai)
and (i, 0)
. Find two lines, which, together with the x-axis forms a container, such that the container contains the most water.
Notice that you may not slant the container.
Example 1:
Input: height = [1,8,6,2,5,4,8,3,7] Output: 49 Explanation: The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.
Example 2:
Input: height = [1,1] Output: 1
Example 3:
Input: height = [4,3,2,1,4] Output: 16
Example 4:
Input: height = [1,2,1] Output: 2
【思路】
木桶短板,能装的最多水,由最短的那条决定
头尾俩指针,计算面积max更新
判断左右哪边更小(木桶更短板),更矮的向对面移动,left<right
因为小的移动,下一个值可能更大,面积也可能更大
【代码】
class Solution { public int maxArea(int[] height) { int area=0; int left=0; int right=height.length-1; while(left<right){ area=Math.max(area,Math.min(height[left],height[right])*(right-left)); if(height[left]<height[right]) left++; else right--; } return area; } }