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 }