• 簡化問題後的代碼

• /*
      原問題：
      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.
  
      解析：這是一道考察算法的題目，如果你沒有get到那個點，就會搞得很復雜，但是簡化了問題之後就會恍然大悟，非常的簡單
      簡化的代碼和原來的代碼如下
   
  */class Solution {
  public:
      int maxArea(vector<int>& height) {
          int left = 0;
          int right = height.size() - 1;
          int max = 0;
          while (right > left)
          {
              max = std::max(min(height[right], height[left]) * (right - left), max);
              if (height[right] > height[left])
                  left
   
  ++;
              else
                  right--;
          }
          return max;
      }
  };

• 原來自己寫的復雜的代碼

• struct structure {
      int height;// 表示高度
      int sub;// 表示下標
      structure(int height, int sub):height(height), sub(sub){}
  };
  class Solution {
  public:
      int maxArea(vector<int>& height) {
          
   
  int max_height = *max_element(height.begin(), height.end());
  
          // left容器存放從左往右遞增的高度，到最高為止
          // right容器存放從有往左遞增的高度，到最高為止
          vector<structure> left(1, structure(height[0], 0));
          vector<structure> right(1, structure(height[height.size() - 1], height.size() - 1));
  
          // 填充left容器
          int index = 1;
          while (true)
          {
              if (left[left.size() - 1].height == max_height)
                  break;
  
              if (height[index] > left[left.size()-1].height)
                  left.insert(left.end(), structure(height[index], index));
              index++;
          }
  
          // 填充right容器
          index = height.size() - 2;
          while (true)
          {
              if (right[right.size() - 1].height == max_height)
                  break;
  
              if (height[index] > right[right.size() - 1].height)
                  right.insert(right.end(), structure(height[index], index));
              index--;            
          }
  
          // 不斷將left、right容器的最小的高度pop出來計算該高度水位的面積，找出最大的面積
          int max_area = 0;
          while (left.size() != 1 || right.size() != 1)
          {
              if (left[0].height < right[0].height)
              {
                  int area = left[0].height * (right[0].sub - left[0].sub);
                  if (area > max_area)
                      max_area = area;
                  left.erase(left.begin());
              }
              else
              {
                  int area = right[0].height * (right[0].sub - left[0].sub);
                  if (area > max_area)
                      max_area = area;
                  right.erase(right.begin());
              }
          }
  
          // 單獨處理left、right容器各自剩下的最後一個max_height
          int area = left[0].height * (right[0].sub - left[0].sub);
          if (area > max_area)
              max_area = area;
  
          return max_area;
      }
  };

• 1234213412

leetcode11 裝最多水的容器（要學會簡化問題）