[Leetcode] Container With Most Water (Java)
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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.
题意就是在坐标轴上找两条线x=l,x=r,使(r-l)*min(height[l],height[r])最大
从两边往中间找,不停找比当前线更长的线,计算面积大小,若比当前最大面积大,则更新,然后继续找,直到两边碰头
public class ContainerWithMostWater {public int maxArea(int[] height) {int lindex = 0;int rindex = height.length-1;int max = (rindex-lindex)*Math.min(height[lindex],height[rindex]);while(rindex>lindex) {if(height[lindex]<height[rindex]) {int k = lindex+1;while(height[k]<height[lindex]&&k<rindex)k++;lindex=k;}else {int k = rindex-1;while(height[k]<height[rindex]&&k>lindex) k--;rindex=k;}int temp = (rindex-lindex)*Math.min(height[rindex],height[lindex]);if(max<temp){max=temp;}}return max;}public static void main(String[] args) {int[] height = {3,2,1,3};System.out.println(new ContainerWithMostWater().maxArea(height));}}
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