[Leetcode] Search a 2D Matrix-白红宇

[Leetcode] Search a 2D Matrix

阅读量：6237 次

发布时间：2019-06-22

本文共 2512 字，大约阅读时间需要 8 分钟。

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.

从右上角向左下角搜索，初始row = 0, col = n - 1; 如果当前值等于目标值，返回真，如果以目标值大，则舍弃当前列，否则舍弃当前行。

1 class Solution { 2 public: 3     bool searchMatrix(vector
     
      
        > &matrix, int target) { 4         int row = 0, col = matrix[0].size() - 1; 5         while (row < matrix.size() && col >=0) { 6             if (matrix[row][col] == target) { 7                 return true; 8             } else if (matrix[row][col] > target) { 9                 --col;10             } else {11                 ++row;12             }13         }14         return false;15     }16 };

第二遍做的时候发现其实这个矩阵不完全是杨氏矩阵，可以使用二分搜索，下面是代码。

1 class Solution { 2 public: 3     bool searchMatrix(vector
     
      
        > &matrix, int target) { 4         int m = matrix.size(); 5         int n = matrix[0].size(); 6         int L = 0, R = n * m - 1, M; 7         int col, row; 8         while (L <= R) { 9             M = L + ((R - L) >> 1);10             row = M / n;11             col = M % n;12             if (matrix[row][col] > target) R = M - 1;13             else if (matrix[row][col] < target) L = M + 1;14             else return true;15         }16         return false;17     }18 };

上面是对所有元素二分，复杂度是log(n*m)，还可以更优化一点，对于普通的杨氏矩阵也适用，复杂度为log(n)+log(m).

1 class Solution { 2 public: 3     bool searchMatrix(vector
     
      
        > &matrix, int target) { 4         int row = lowerBound(matrix, target); 5         if (row == -1) return false; 6         return binSearch(matrix, target, row); 7     } 8      9     int lowerBound(vector
       
        
          > &matrix, int target) {10         int L = 0, R = matrix.size() - 1, M;11         while (L <= R) {12             M = L + ((R - L) >> 1);13             if (matrix[M][0] <= target) L = M + 1;14             else R = M - 1;15         }16         return R;17     }18     19     bool binSearch(vector
         
          
            > &matrix, int target, int row) {20 int L = 0, R = matrix[row].size() - 1, M;21 while (L <= R) {22 M = L + ((R - L) >> 1);23 if (matrix[row][M] < target) L = M + 1;24 else if (matrix[row][M] > target) R = M - 1;25 else return true;26 }27 return false;28 }29 };