leetcode 11. Container With Most Water

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.

class Solution {
public:
	int maxArea(vector<int>& height) {
		int heightSize = height.size();
		if (height.empty())
			return 0;
		if (heightSize < 2)
			return 0;
		int index=0;
		for (int i = 0; i < height.size(); i++)
			if (height[i]>height[index])
				index= i;
		if (height[index] == 0)
			return 0;
		int max = 0;
		vector<int>starter(heightSize);
		if (index < height.size() / 2)
		{
			int nxt = 0; int i = 0;
			bool f = false;
			while (!f&&i < heightSize - 1)
			{
				if (height[i] == 0)
					i++;
				else
				{
					f = true;
					i = nxt;
					nxt = heightSize;
					if (starter[i] == 0)
					{
						for (int j = i + 1; j < heightSize; j++)
						{
							if (height[j] <= height[i])
								starter[j] = 1;
							else
							{
								if (f)
								{
									nxt = j;
									f = false;
								}
							}
							int h = height[i] < height[j] ? height[i] : height[j];
							if (h*(j - i)>max)
								max = h*(j - i);
						}
					}
				}
			}
		}
		else
		{
			int nxt = heightSize - 1; int i = heightSize - 1;
			bool f = false;
			while (!f&&i >= 1)
			{
				if (height[i] == 0)
					i--;
				else
				{
					f = true;
					i = nxt;
					nxt = 0;
					if (starter[i] == 0 && height[i] > 0)
					{
						for (int j = i - 1; j >= 0; j--)
						{
							if (height[j] <= height[i])
								starter[j] = 1;
							else
							{
								if (f)
								{
									nxt = j;
									f = false;
								}
							}
							int h = height[i] < height[j] ? height[i] : height[j];
							if (h*(i - j)>max)
								max = h*(i - j);
						}
					}
				}
			}
		}
		return max;
	}
};

accept