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
.
思路:先利用每一行的最后一个数与target判断,确定target可能在的行数;再在确定的那一行进行折半查找。
public class S074 { public boolean searchMatrix(int[][] matrix, int target) { int i = 0; for (;i<matrix.length;i++) { if (matrix[i][matrix[0].length-1]>=target) { break; } } if (i == matrix.length) return false; int l = 0, r = matrix[0].length-1; //折半查找 while (l <= r) { if (target == matrix[i][(l+r)/2]) return true; else if (target > matrix[i][(l+r)/2]) l = (l+r)/2+1; else r = (l+r)/2-1; } return false; } }
时间: 2024-11-03 05:38:24