[LeetCode] 011. Container With Most Water (Medium) (C++/Java/Python)

索引：[LeetCode] Leetcode 题解索引 (C++/Java/Python/Sql)

Github: https://github.com/illuz/leetcode

011.Container_With_Most_Water (Medium)

链接：

题目：https://oj.leetcode.com/problems/container-with-most-water/

代码(github)：https://github.com/illuz/leetcode

题意：

给一些挡板，选两个挡板，求最大蓄水容量。

分析：

可以看看这个大神的详细算法，给跪…

1. 暴力 O(n*n) 会超时
2. 双指针，O(n) 时间和 O(1) 空间，应该是最优的算法了，上述的文章有这个算法的正确性证明。
3. 预处理每个挡板的左边最高和右边最高，这样蓄水区间就可以知道了

这里只用了第二种算法。

代码：

C++:

class Solution {
public:
    int maxArea(vector<int> &height) {
		int lpoint = 0, rpoint = height.size() - 1;
		int area = 0;
		while (lpoint < rpoint) {
			area = max(area, min(height[lpoint], height[rpoint]) *
					(rpoint - lpoint));
			if (height[lpoint] > height[rpoint])
				rpoint--;
			else
				lpoint++;
		}
		return area;
    }
};

Java:

public class Solution {

    public int maxArea(int[] height) {
        int lpoint = 0, rpoint = height.length - 1;
        int area = 0;
        while (lpoint < rpoint) {
            area = Math.max(area, Math.min(height[lpoint], height[rpoint]) *
                    (rpoint - lpoint));
            if (height[lpoint] > height[rpoint])
                rpoint--;
            else
                lpoint++;
        }
        return area;
    }
}

Python:

class Solution:
    # @return an integer
    def maxArea(self, height):
        lp, rp = 0, len(height) - 1
        area = 0
        while lp < rp:
            area = max(area, min(height[lp], height[rp]) * (rp - lp))
            if height[lp] > height[rp]:
                rp -= 1
            else:
                lp += 1
        return area