[LeetCode] 11. Container With Most Water 装最多水的容器

Given n non-negative integers a₁, a₂, ..., a_n, where each represents a point at coordinate (i, a_i). n vertical lines are drawn such that the two endpoints of line i is at (i, a_i) 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

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