Container With Most Water "最多盛水问题"
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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 and n is at least 2.
测试样例:1 8 6 2 5 4 8 3 7
输出:49
解释:对于给定数组,盛水容器的宽度为数组中两个元素坐标差(假设为r-l),高度为两个元素中的最小值(假设为Min(height[l]、height[r])),注意输入无序且非负,数量级为10^7.
思路一:穷举
复杂度:O(n^2),超时
思路二:“两头堵”
容器容量的大小由高度与宽度共同决定,从宽度最大开始判断,容量由受限于高度最小者Min(height[l]、height[r]),所以将最小高度的坐标单向移动,以高度补偿宽度的损失。记录下过程中出现的最大容量,当左右坐标(l、r)重合时,遍历结束。
示意图如下,图片转自Leedcode,链接:https://leetcode.com/problems/container-with-most-water/solution/
class Solution { public int maxArea(int[] height) { int result = 0; int l = 0; int r = height.length-1; while(l<r){//左右坐标未相遇。当相遇时即完成了所有坐标位置的检查,且结果是单调递增。 result = Math.max(result,(r-l)*Math.min(height[l],height[r]));//计算当前容量,计算下历史与当前中的最大值 if(height[l]>height[r]){//标胶左右坐标,调整矮的那个,以高度补偿宽度损失 r--; }else{ l++; } } return result; }}
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