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 and n is at least 2.

The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.

Example:

Input: [1,8,6,2,5,4,8,3,7]
Output: 49

简单来说就是给一堆高，问取哪两个高围的面积最大

思路1：暴力

没什么好说的 直接过一遍

#define MIN(A,B) A<B?A:B
int maxArea1(vector<int>& height) {
	int sz = height.size(), res = 0, temp = 0;
	for(int i=0;i<sz;++i)
		for (int j = i + 1; j < sz; ++j) {
			temp = (j - i)*(MIN(height[i], height[j]));
			if (temp > res) res = temp;
		}
	return res;
}

思路2：优化

在上述暴力方法的基础上，寻找可以优化的地方
假设初始面积是取最左和最右两个高计算得到，从两边向中间扫描

每次选当前两高中较短的一边向另一边移动，计算面积并与之前值比较。
这样操作时因为，边移动时会使两边距离变小，而计算面积时是根据较短边的高度计算的，移动较长边不会得到任何收益（移动到更长边还是使用另一边高，移动到更短边会使面积再次减少）。

    #define MIN(A,B) A<B?A:B
	int maxArea(vector<int>& height) {
		int j = height.size()-1, res = 0, temp = 0;
		int i = 0;
		while(i<j){
			temp = (j - i)*(MIN(height[i], height[j]));
			if (temp > res) res = temp;
			height[i] > height[j] ? --j : ++i;
		}
		return res;
	}