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
.
1.迭代
class Solution { public: bool searchMatrix(vector<vector<int>>& matrix, int target) { int m = matrix.size(); int n = m==0 ? 0:matrix[0].size(); int i=0,j=n-1; while(i<m && j>=0){ if(matrix[i][j] < target) i++; else if(matrix[i][j] > target) j--; else return true; } return false; } };
2.递归
class Solution { public: bool searchMatrix(vector<vector<int>>& matrix,int r,int c,int target){ if(c<0 || r >= matrix.size()){ return false; } if(matrix[r][c]>target){ return searchMatrix(matrix,r,c-1,target); }else if(matrix[r][c]<target){ return searchMatrix(matrix,r+1,c,target); }else{ return true; } } bool searchMatrix(vector<vector<int>>& matrix, int target) { int m = matrix.size(); if(m==0){ return false; } int n = matrix[0].size(); return searchMatrix(matrix,0,n-1,target); } };