11. Container With Most Water
Description
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.
Example:
Input: [1,8,6,2,5,4,8,3,7]
Output: 49
Solution: (Java)
class Solution {
public int maxArea(int[] height) {
int most = 0, container;
for (int i = 0; i < height.length - 1; i++) {
for (int j = i + 1; j < height.length; j++) {
container = (j-i)*(Math.min(height[j], height[i]));
if (container > most) {
most = container;
}
}
}
return most;
}
}
思路:
这里直接用两个 for 循环,逐次计算区域面积,保存最大的即可,时间复杂度为 O(n2),accept 之后看了其他的思路,即下面方法2的思路:用头尾两个指针,对应高度较小的指针往中间缩减(只由较小的长度决定面积),依次计算区域面积,时间复杂度为 O(n).
方法2:
public int maxArea2(int[] height) {
int most = 0, left = 0, right = height.length - 1;
while (left < right) {
most = Math.max(most, (right-left) * (Math.min(height[left], height[right])));
if (height[left] < height[right])
left++;
else
right--;
}
return most;
}