改进此页

Search a 2D Matrix

Source

Problem

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.

Example

Consider the following matrix:

[
    [1, 3, 5, 7],
    [10, 11, 16, 20],
    [23, 30, 34, 50]
]

Given target = 3, return true.

Challenge

O(log(n) + log(m)) time

题解 - 一次二分搜索 V.S. 两次二分搜索

  • 一次二分搜索 - 由于矩阵按升序排列，因此可将二维矩阵转换为一维问题。对原始的二分搜索进行适当改变即可(求行和列)。时间复杂度为 O(log(mn))=O(log(m)+log(n))O(log(mn))=O(log(m)+log(n))
  • 两次二分搜索 - 先按行再按列进行搜索，即两次二分搜索。时间复杂度相同。

显然我们应该选择一次二分搜索，直接上 lower bound 二分模板。

Java

public class Solution {
    /**
     * @param matrix, a list of lists of integers
     * @param target, an integer
     * @return a boolean, indicate whether matrix contains target
     */
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0 || matrix[0] == null) {
            return false;
        }

        int ROW = matrix.length, COL = matrix[0].length;
        int lb = -1, ub = ROW * COL;
        while (lb + 1 < ub) {
            int mid = lb + (ub - lb) / 2;
            if (matrix[mid / COL][mid % COL] < target) {
                lb = mid;
            } else {
                if (matrix[mid / COL][mid % COL] == target) {
                    return true;
                }
                ub = mid;
            }
        }

        return false;
    }
}

源码分析

仍然可以使用经典的二分搜索模板(lower bound)，注意下标的赋值即可。

  1. 首先对输入做异常处理，不仅要考虑到matrix为null，还要考虑到matrix[0]的长度也为0。
  2. 由于 lb 的变化处一定小于 target, 故在 else 中判断。

复杂度分析

二分搜索，O(logn)O(\log n).