Search a 2D Matrix

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 public class Solution {
 2     public boolean searchMatrix(int[][] matrix, int target) {
 3         // Start typing your Java solution below
 4         // DO NOT write main() function
 5         if(matrix == null || matrix.length == 0 || matrix[0].length == 0) return false;
 6         int i = 0;
 7         for(; i < matrix.length; i ++){
 8             if(target < matrix[i][0]) {
 9                 if(i < 1) return false;
10                 for(int j = 0; j < matrix[i-1].length; j ++){
11                     if(target == matrix[i-1][j]) return true;
12                 }
13                 return false;
14             }
15             else if(target == matrix[i][0]) return true;
16         }
17         for(int j = 0; j < matrix[i-1].length; j ++){
18             if(target == matrix[i-1][j]) return true;
19         }
20         return false;
21     }
22 }

 第二遍：使用两次二分来加速。 一次是找到不大于target的col= 0的最值， 第二遍遍历这一row 然后找target。 如果找不到 就return false。

 1 public class Solution {
 2     public boolean searchMatrix(int[][] matrix, int target) {
 3         // Start typing your Java solution below
 4         // DO NOT write main() function
 5         if(matrix == null || matrix.length == 0 || matrix[0].length == 0) return false;
 6         int start = 0, end = matrix.length - 1, mid = 0;
 7         int i = 0, j = 0;
 8         while(start <= end){
 9             mid = (start + end) / 2;
10             int tmp = matrix[mid][0];
11             if(tmp < target){
12                 i = mid;
13                 start = mid + 1;
14             }else if(tmp == target) return true;
15             else{
16                 end = mid - 1;
17             }
18         }
19         start = 0;
20         end = matrix[0].length - 1;
21         while(start <= end){
22             mid = (start + end) / 2;
23             int tmp = matrix[i][mid];
24             if(tmp < target){
25                 j = mid;
26                 start = mid + 1;
27             }else if(tmp == target) return true;
28             else{
29                 end = mid - 1;
30             } 
31         }
32         return false;
33     }
34 }

 Step的方法：

 1 public class Solution {
 2     public boolean searchMatrix(int[][] matrix, int target) {
 3         // Start typing your Java solution below
 4         // DO NOT write main() function
 5         if(matrix == null || matrix.length == 0 || matrix[0].length == 0) return false;
 6         int row = matrix.length - 1, col = 0;
 7         while(row > -1 && col < matrix[0].length){
 8             if(matrix[row][col] > target) row --;
 9             else if(matrix[row][col] < target) col ++;
10             else return true;
11         }
12         return false;
13     }
14 }

 

posted on 2013-10-01 14:25  Step-BY-Step  阅读(163)  评论(0)  编辑  收藏  举报

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