[LeetCode] 74. Search a 2D Matrix 解题思路

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

• Integers in each row are sorted from left to right.
• The first integer of each row is greater than the last integer of the previous row.

For example,

Consider the following matrix:

[
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]

Given target = 3, return true.

问题：在给定的已排序的二维矩阵中，判断是否包含某个整数。

思路：看到二维数组，首先想到的是二维数组和 一维数组可以直接转换，arr[i][j] 等于 arr[i*col + j]。而在一个已排序的一维数组中搜索一个元素，可以采用分治(Divide and Conquer)思想，更具体些就是二分搜索(Binary Search) 算法。

将上面的结合起来就是，先实现一维数组的二分搜索算法，然后将算法中的一位转换为二维数组即可。

 bool searchMatrix(vector<vector<int>>& matrix, int target) {

     if (matrix.size() == ) {
         return false;
     }

     int col = (int)matrix[].size();
     int row = (int)matrix.size();

     if (col ==  && row == ) {
         return ( matrix[][] == target );
     }

     int lm = ;
     int rm = col * row - ;

     while (lm < rm ) {

         if (lm +  == rm) {
             int quoL = lm / col;
             int remL = lm % col;

             int quoR = rm / col;
             int remR = rm % col;

             if (matrix[quoL][remL] == target || matrix[quoR][remR] == target) {
                 return true;
             }else{
                 return false;
             }
         }

         int midm = (lm + rm) / ;

         int quo = midm / col;
         int rem = midm % col;

         if (matrix[quo][rem] == target) {
             return true;
         }

         if (matrix[quo][rem] < target) {
             lm = midm;
         }else{
             rm = midm;
         }
     }

     return false;
 }