题目:Container With Most Water 装最多水的容器
难度:中等
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.
The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.
Example:
Input: [1,8,6,2,5,4,8,3,7] Output: 49
题意解析:
给定 n 个非负整数 a1,a2,...,an,每个数代表坐标中的一个点 (i, ai) 。在坐标内画 n 条垂直线,垂直线 i 的两个端点分别为 (i, ai) 和 (i, 0)。找出其中的两条线,使得它们与 x 轴共同构成的容器可以容纳最多的水。
说明:你不能倾斜容器,且 n 的值至少为 2。
图中垂直线代表输入数组 [1,8,6,2,5,4,8,3,7]。在此情况下,容器能够容纳水(表示为蓝色部分)的最大值为 49。
解题思路一:
对于给定的数组,我们希望长度尽可能的长,高度则只能根据两个之间小的那个来计算,因此,我们从数组的两端开始向中间缩进。判断当前的容积,然后将小的向右(左小)或向左(右小)移动即可。
public int maxArea(int[] height) {
int i=0,j = height.length-1,max = 0;
while (i < j){
max = Math.max(max, Math.min(height[i], height[j]) * (j - i));
if (height[i] < height[j]){
i++;
}else {
j--;
}
}
return max;
}
此算法时间复杂度O(n)
提交此代码之后:
Runtime: 2 ms, faster than 99.16% of Java online submissions for Container With Most Water.
Memory Usage: 40.8 MB, less than 8.02% of Java online submissions for Container With Most Water.