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
.
Solution1:
分两次2分查找,先定位哪一行,然后再在那一行里寻找目标。
public boolean searchMatrix(int[][] matrix, int target) { int m = matrix.length, n = matrix[0].length; int low = 0, high = m-1; while(low <= high) { int mid = (low+high) / 2; if(matrix[mid][0] == target) { return true; } else if(matrix[mid][0] < target) { low++; } else { high--; } } int row = low; if(row>=m || (matrix[row][0]>target && row > 0)) { row--; } low = 0; high = n-1; while(low <= high) { int mid = (low+high) / 2; if(matrix[row][mid] == target) { return true; } else if(matrix[row][mid] < target) { low++; } else { high--; } } return false; }
Solution2:
把二维数组的坐标转化成一位数组,整体进行二分查找。实现更简单。
public boolean searchMatrix(int[][] matrix, int target) { int m = matrix.length, n = matrix[0].length; int low = 0, high = m*n-1; while(low <= high) { int mid = (low+high) / 2; int row = mid / n; int col = mid % n; int num = matrix[row][col]; if(num == target) { return true; } else if(num < target) { low = mid + 1; } else { high = mid - 1; } } return false; }
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