【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;

    }

};

巴特西

【Container With Most Water】cpp

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