标签:
Container With Most Water
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.
思路:
双指针,减少短板
我的代码:
public class Solution { public int maxArea(int[] height) { if(height == null || height.length <= 1) return 0; int left = 0; int right = height.length - 1; int maxArea = Math.min(height[left],height[right])*(right - left); while(left < right) { int curLeft = left; int curRight = right; if(height[left] < height[right]) { while(left < right && height[left] <= height[curLeft]) { left++; } maxArea = Math.max(Math.min(height[left],height[right])*(right - left), maxArea); } else { while(left < right && height[right] <= height[curRight]) { right--; } maxArea = Math.max(Math.min(height[left],height[right])*(right - left), maxArea); } } return maxArea; } }
他人代码:
public class Solution { public int maxArea(int[] height) { if (height == null) { return 0; } int left = 0; int right = height.length - 1; int maxArea = 0; while (left < right) { int h = Math.min(height[left], height[right]); int area = h * (right - left); maxArea = Math.max(maxArea, area); if (height[left] < height[right]) { // 如果左边界比较低,尝试向右寻找更高的边界 left++; } else { // 如果右边界比较低,尝试向左寻找更高的边界 right--; } } return maxArea; } }
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原文地址:http://www.cnblogs.com/sunshisonghit/p/4322150.html