【Container With Most Water】cpp

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.

代码：

class Solution {
public:
    int maxArea(vector<int>& height) {
            if ( height.size()< ) return ;
            int max_area = ;
            int left = ;
            int right = height.size()-;
            while ( left<right )
            {
                if ( height[left]<=height[right] )
                {
                    max_area = std::max(max_area, (right-left)*height[left]);
                    left++;
                }
                else
                {
                    max_area =  std::max(max_area, (right-left)*height[right]);
                    right--;
                }
            }
            return max_area;
    }
};

tips：

试图用DP去做，但是没想出来；最后无奈落入了Greedy的俗套solution。

这个greedy的思路也是蛮巧的：从两头开始往中间greedy，头尾两个greedy一起变化才得到greedy的条件。

=======================================

第二次过这道题，这道题题意不清晰：如果选了某两个板子，就当其他板子不存在。

class Solution {
public:
    int maxArea(vector<int>& height) {
            int l = ;
            int r = height.size()-;
            int ret = ;
            while ( l<r )
            {
                if ( height[l]<=height[r] )
                {
                    ret = max(ret, (r-l)*height[l]);
                    l++;
                }
                else
                {
                    ret = max(ret, (r-l)*height[r]);
                    r--;
                }
            }
            return ret;
    }
};