LeetCode – Search a 2D Matrix II (Java)
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.
Java Solution 1
In a naive approach, we can use the matrix boundary to reduce the search space. Here is a simple recursive implementation.
public boolean searchMatrix(int[][] matrix, int target) { int i1=0; int i2=matrix.length-1; int j1=0; int j2=matrix[0].length-1; return helper(matrix, i1, i2, j1, j2, target); } public boolean helper(int[][] matrix, int i1, int i2, int j1, int j2, int target){ if(i1>i2||j1>j2) return false; for(int j=j1;j<=j2;j++){ if(target < matrix[i1][j]){ return helper(matrix, i1, i2, j1, j-1, target); }else if(target == matrix[i1][j]){ return true; } } for(int i=i1;i<=i2;i++){ if(target < matrix[i][j1]){ return helper(matrix, i1, i-1, j1, j2, target); }else if(target == matrix[i][j1]){ return true; } } for(int j=j1;j<=j2;j++){ if(target > matrix[i2][j]){ return helper(matrix, i1, i2, j+1, j2, target); }else if(target == matrix[i2][j]){ return true; } } for(int i=i1;i<=i2;i++){ if(target > matrix[i][j2]){ return helper(matrix, i1, i+1, j1, j2, target); }else if(target == matrix[i][j2]){ return true; } } return false; } |
Java Solution 2
Time Complexity: O(m + n)
public boolean searchMatrix(int[][] matrix, int target) { int m=matrix.length-1; int n=matrix[0].length-1; int i=m; int j=0; while(i>=0 && j<=n){ if(target < matrix[i][j]){ i--; }else if(target > matrix[i][j]){ j++; }else{ return true; } } return false; } |
<pre><code> String foo = "bar"; </code></pre>
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alexwest11
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Brenda Martis