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 in ascending from left to right.
- Integers in each column are sorted in ascending from top to bottom.
For example,
Consider the following matrix:
[ [1, 4, 7, 11, 15], [2, 5, 8, 12, 19], [3, 6, 9, 16, 22], [10, 13, 14, 17, 24], [18, 21, 23, 26, 30] ]
Given target = 5, return true.
Given target = 20, return false.
本题目，开始的时候想法是遍历每一行，每一行使用binary search来做，时间复杂度是O(mlongn)。代码如下：
1 public class Solution { 2 public boolean searchMatrix(int[][] matrix, int target) { 3 if(matrix==null||matrix.length==0||matrix[0].length==0) return false; 4 int m = matrix.length; 5 int n = matrix[0].length; 6 for(int i=0;i<m;i++){ 7 int left = 0,right = n-1; 8 while(left<=right){ 9 int mid = left+(right-left)/2; 10 if(matrix[i][mid]==target) return true; 11 else if(matrix[i][mid]<target) left = mid+1; 12 else right = mid-1; 13 } 14 } 15 return false; 16 } 17 } 18 //the run time complexity could be O(mlongn), the space complexity could be O(1);
看了答案才知道原来可以达到O(m+n)来做，代码如下：
1 public class Solution { 2 public boolean searchMatrix(int[][] matrix, int target) { 3 if(matrix==null||matrix.length==0||matrix[0].length==0) return false; 4 int m = matrix.length; 5 int n = matrix[0].length; 6 int row = 0; 7 int col = n-1; 8 while(row<m&&col>=0){ 9 if(matrix[row][col]==target) return true; 10 else if(matrix[row][col]<target&&row<m-1){ 11 row++; 12 }else if(matrix[row][col]>target&&col>0){ 13 col--; 14 }else{ 15 break; 16 } 17 } 18 return false; 19 } 20 } 21 //the run time complexity would be O(m+n),the space complexity could be O(1);