leetcode11 装最多水的容器（要学会简化问题）

标签：vector   高度   lines   nta   least   原来   就会   处理   ++   

•  简化问题后的代码
•  

  /*
      原问题：
      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：这是一道考察算法的题目，如果你没有get到那个点，就会搞得很复杂，但是简化了问题之后就会恍然大悟，非常的简单
      简化的代码和原来的代码如下
  */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;
      }
  };

   

• 原来自己写的复杂的代码

• struct structure {
      int height;// 表示高度
      int sub;// 表示下标
      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容器存放从左往右递增的高度，到最高为止
          // right容器存放从有往左递增的高度，到最高为止
          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--;            
          }
  
          // 不断将left、right容器的最小的高度pop出来计算该高度水位的面积，找出最大的面积
          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());
              }
          }
  
          // 单独处理left、right容器各自剩下的最后一个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 装最多水的容器（要学会简化问题）

标签：vector   高度   lines   nta   least   原来   就会   处理   ++   

原文地址：https://www.cnblogs.com/jkn1234/p/8969744.html