Container With Most Water

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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.

思路：從兩側往中間遍曆。使用curHeight儲存當前計算過的容器的最大高度。由於遍曆是從兩側往中間進行的，因此，這將保證任何高度不大於curHeight的容器的最大容量已經被計算過。於是，我們只需在當前兩端的高度都大於curHeight時，才更新curHeight，同時計算以新高度為高度的容器的最大容量。

 1 class Solution { 2 public: 3     int maxArea( vector<int> &height ) { 4         int area = 0, curHeight = 0; 5         int left = 0, right = (int)height.size()-1; 6         while( left < right ) { 7             if( height[left] <= curHeight ) { ++left; continue; } 8             if( height[right] <= curHeight ) { --right; continue; } 9             curHeight = min( height[left], height[right] );10             area = max( area, (right-left)*curHeight );11         }12         return area;13     }14 };

 

Container With Most Water