leetcode 11:Container With Most Water
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【题目】
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.
public class Solution {
public int maxArea(int[] height) {
}
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.
我选择使用Java语言
public class Solution {
public int maxArea(int[] height) {
}
}
【解析】题意:在二维坐标系中,(i, ai) 表示 从 (i, 0) 到 (i, ai) 的一条线段,任意两条这样的线段和 x 轴组成一个木桶,找出能够盛水最多的木桶,返回其容积。
做法:使用left和right分别标记最左边的木板和最右边的木板,试想,当height[left]<=height[right]时计算容积为height[left]*(right-left)。此时如果将right左移的话,容积不会增加,我们要抛弃计算它们。计算其它可能性是只能将left++。height[left]>height[right]时分析类似,应该right--。最终循环在left==right时停止。
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