[LeetCode] 011. Container With Most Water (Medium) (C++/Java/Python)
2015-03-02 22:51
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索引:[LeetCode] Leetcode 题解索引 (C++/Java/Python/Sql)
Github: https://github.com/illuz/leetcode
代码(github):https://github.com/illuz/leetcode
暴力 O(n*n) 会超时
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。
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。
Java:
Python:
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…暴力 O(n*n) 会超时
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。
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
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