217. Container With Most Water
You are given an integer array height of length n. There are n vertical lines drawn such that the two endpoints of the ith line are (i, 0) and (i, height[i]).
Find two lines that together with the x-axis form a container, such that the container contains the most water.
Return the maximum amount of water a container can store.
Notice that you may not slant the container.
Constraints:
- n == height.length
- 2 <= n <= 10^5
- 0 <= height[i] <= 10^4
Example
Input: height = [1,8,6,2,5,4,8,3,7]
Output: 49
Explanation: 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.
思路
求最大面积,那影响面积的两个条件就是长和宽,宽就是数组的值,长就是元素之间相差的位数
其中决定宽的应该是一对值中较小的那个值,大的那个值能适配的肯定更多些,所以就移动小的那一位
题解
- 无脑快速AC
public int maxArea(int[] height) {
int result, left, right, step;
result = left = 0;
right = height.length - 1;
while (left < right) {
// 找宽
int len = Math.min(height[left], height[right]);
// 算长
step = right - left;
result = Math.max(len * step, result);
// 这里要把移位放最后做,不然会直接跳过第一位,其中也会有左边等于右边的情况,这里就让左边移动了
if (height[left] > height[right])
right--;
else
left++;
}
return result;
}
标签:11,right,container,int,height,Water,Most,result,left
From: https://www.cnblogs.com/tanhaoo/p/17041184.html