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.
与42. Trapping Rain Water 类似，两条垂直的线和X轴组成一个容器，灌水多少不仅与两个柱子的高度有关，也与两个柱子的距离有关，公式：S(i,j) = min(ai, aj) * (j-i)，容器不能倾斜，求容纳最多水的两个线组合。
用暴力搜索Brute Force， Time: O(n2)会超时。
使用双指针two pointers。 定义left，right两个指针，两指针相遇循环结束。何时移动左右指针呢，保留较高的柱子，移动较矮的柱子。
Time complexity: O(n), Space complexity: O(1)
Java:
class Solution {
public int maxArea(int[] height) {
int l = 0;
int r = height.length - 1;
int ans = 0;
while (l < r) {
int h = Math.min(height[l], height[r]);
ans = Math.max(ans, h * (r - l));
if (height[l] < height[r])
++l;
else
--r;
}
return ans;
}
}
Python:
class Solution:
def maxArea(self, height):
max_area, i, j = 0, 0, len(height) - 1
while i < j:
max_area = max(max_area, min(height[i], height[j]) * (j - i))
if height[i] < height[j]:
i += 1
else:
j -= 1
return max_area
C++:
class Solution {
public:
int maxArea(const vector<int>& height) {
int ans = 0;
int l = 0;
int r = height.size() - 1;
while (l < r) {
int h = min(height[l], height[r]);
ans = max(ans, h * (r - l));
if (height[l] < height[r])
++l;
else
--r;
}
return ans;
}
};
类似题目：
[LeetCode] 42. Trapping Rain Water 收集雨水
