leetcode.11---------Container With Most Water

标签：container with most   算法   acm   leetcode   

  题目：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.

题目的意思大概就是：x轴上在1,2,...,n点上有许多垂直的线段，长度依次是a₁, a₂, ..., an。找出两条线段，使他们和x抽围成的面积最大。面积公式是 Min(a_i, a_j) X |j - i|

方法一：暴力穷举方法

class Solution {
public:
	int maxArea(vector<int> &height)
	{
		int n = height.size();
		int max = 0;
		for (int i = 0; i < n; i++)
		{
			for (int j = i + 1; j < n; j++)
			{
				int temp = (j - i)*(height[i]>height[j] ? height[j] : height[i]);
				if (temp > max)
				{
					max = temp;
					continue;
				}
			}
		}
		return max;
	}
};

The basic idea is simple: use two pointers to indicate the left edge and right edge of the water container. Every time move the lower edge towards another edge with one step. At last, we can find the max water container.

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

改进：

int maxArea(vector<int> &height) {
    int left = 0, right = height.size() - 1;
    int maxWater = 0;
    while(left < right){
        maxWater = max(maxWater, (right - left) * min(height[left], height[right]));
        height[left] < height[right] ? left++ : right--;
    }
    return maxWater;
}

leetcode.11---------Container With Most Water

标签：container with most   算法   acm   leetcode   

原文地址：http://blog.csdn.net/chenxun_2010/article/details/43272467