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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.
思路:可以从最外侧开始考虑,第1层以及第n层所形成的容器当做最初始的容器。每个容器的面积,取决于最短的的木板以及宽度,要想减少宽度之后,能够得到更大的容量,必须保证木板的长度增大了。采用贪心策略可以解决。
class Solution { public: int maxArea(vector<int> &height) { int width = height.size(); if(width <= 1) return 0; int start = 0; int end = width - 1; int result = INT_MIN; while(start < end) { int tempArea = min(height[start], height[end]) * (end - start); result = max(result, tempArea); if(height[start] <= height[end]) start++; else end--; } return result; } };
【Leetcode】Container With Most Water,布布扣,bubuko.com
【Leetcode】Container With Most Water
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原文地址:http://blog.csdn.net/lipantechblog/article/details/31373643