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
至今无法理解怎么想出这种方法的.
从两端向中间逼近. 面积=宽(j-i) x 高
接下来选择i,j中比较矮的向中间靠近即可.
class Solution { public: int maxArea(vector<int>& height) { int res=0; for(int i=0,j=height.size()-1;i<j;) { res=max(res,min(height[i],height[j])*(j-i)); if(height[i]<height[j])++i; else --j; } return res; } };