LintCode-28（Search a 2D Matrix）

關於

lintcode系列，第28題，題目網址：https://www.lintcode.com/problem/search-a-2d-matrix/description

描述

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.

樣例：

樣例  1:
	輸入: [[5]],2
	輸出: false
	
	樣例解釋: 
  沒有包含，返回false。

樣例 2:
	輸入:  
[
  [1, 3, 5, 7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
],3
	輸出: true
	
	樣例解釋: 
	包含則返回true。

思路

二分法的使用個，注意矩陣下標的轉換。

C++實現

class Solution {
public:
    /**
     * @param matrix: matrix, a list of lists of integers
     * @param target: An integer
     * @return: a boolean, indicate whether matrix contains target
     */
    bool searchMatrix(vector<vector<int>> &matrix, int target) {
        // write your code here
      int m=matrix.size();
      if(m == 0) {
        return false;
      }
      int n=matrix[0].size();
      int end=m*n,head=0;
      while(head!=end) {
        if(target == matrix[((head+end)/2)/n][((head+end)/2)%n]) {
          return true;
        }
        else if(target < matrix[((head+end)/2)/n][((head+end)/2)%n]) {
          end = (head+end)/2;
        }
        else {
          head = (head+end)/2 + 1;
        }
      }
      if(matrix[(head/2)/n][(head/2)%n] == target) return true;
      else return false;
      
    }
};