No.011:Container With Most Water
2016-10-09 10:07
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问题:
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.
官方难度:
Medium
翻译:
给定n个非负整数a1,a2,...,an,每一个数值代表坐标轴上的坐标(i,ai)。
画上n条垂直于横坐标的竖线,用于连接点(i,ai)和(i,0)。找到两条线,与x轴一起形成一个容器,能够容纳最多的水。
注意容器不能倾斜。
方法一:
- 利用一个二次循环,同时维护一个最大面积max。
方法一的解题代码:
1 public static int maxArea(int[] height) { 2 if (height == null || height.length < 2) { 3 throw new IllegalArgumentException("Input error"); 4 } 5 // 初始面积 6 int left = 0, right = height.length - 1; 7 int area = (right - left) * Math.min(height[left], height[right]); 8 // 左右侧开始的最长高度 9 int leftMax = height[left]; 10 int rightMax = height[right]; 11 // 从两侧向中间夹逼,即底在不断变小 12 while (left < right) { 13 if (height[left] < height[right]) { 14 left++; 15 // 更小的高没有比较价值 16 if (height[left] <= leftMax) { 17 continue; 18 } else { 19 leftMax = height[left]; 20 } 21 area = Math.max(area, (right - left) * Math.min(height[left], height[right])); 22 } else { 23 right--; 24 if (height[right] <= rightMax) { 25 continue; 26 } else { 27 rightMax = height[right]; 28 } 29 area = Math.max(area, (right - left) * Math.min(height[left], height[right])); 30 } 31 } 32 return area; 33 }maxArea
相关链接:
https://leetcode.com/problems/container-with-most-water/
PS:如有不正确或提高效率的方法,欢迎留言,谢谢!
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