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.
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:
Input: [1,8,6,2,5,4,8,3,7] Output: 49题意：
给定一堆高矮不一的平行的墙，选择其中两堵作为两壁，问最多能存多少水。
Solution1：
Two Pointers: set pointer left at index 0; set pointer right at index len - 1
We calculate area base on width (gap of left and right index ) and height(smaller height matters because of "Short-board effect" )
In order to find max area, we move the pointer whichever representing the smaller height
code:
1 /* 2 Time complexity : O(n). We traverse the whole give array 3 Space complexity : O(1). We only used constant extra space. 4 */ 5 6 class Solution { 7 public int maxArea(int[] height) { 8 int left = 0; 9 int right = height.length - 1; 10 int area = Integer.MIN_VALUE; 11 while(left < right){ 12 area = Math.max(area, Math.min(height[left], height[right]) * (right - left)); 13 if(height[left] < height[right]){ 14 left ++; 15 }else{ 16 right --; 17 } 18 } 19 return area; 20 } 21 }
转载于:https://www.cnblogs.com/liuliu5151/p/10657814.html
