[LintCode] Search a 2D Matrix

Search a 2D Matrix

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

SOLUTION:

这题看到排序数组,就有可能用二分法,然后看到challenge是logn + logm的,就肯定是二分了,第一眼看上去是应该两个二分,找到在哪一行,再在这一行里找,但是,仔细看题目,这个matrix是绝对递增的,就是第二行第一个数绝对大于迪一行最后一个数,也就是说,把每一行都摆在一起,就是一个连续的排列的递增的数组。。。这就好办了,一个binary search就解决了。

然后就是怎么把matrix表示成一行,怎么表示end。

end 可以等于 (行数*列数 -1),然后对应的A[mid] = matrix[mid/行数] [mid%行数] : 第一个行数,就是直接用除法,看有几个行就是在第几行;然后mid去%一下,就是看它在这一行的什么位置。

最后看代码:

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){
            return false;
        }
        if (matrix[0] == null || matrix[0].length == 0){
            return false;
        }
        int row = matrix.length;
        int column = matrix[0].length;
        int start = 0;
        int end = matrix.length * matrix[0].length - 1;
        while (start + 1 < end){
            int mid = start + (end - start) / 2;
            if (matrix[mid / column][mid % column] == target){
                return true;
            }
            if (matrix[mid / column][mid % column] > target){
                end = mid;
            }
            if (matrix[mid / column][mid % column] < target){
                start = mid;
            }
        }
        if (matrix[start / column][start % column] == target){
            return true;
        }
        if (matrix[end / column][end % column] == target){
            return true;
        }
        return false;
    }
}

时间: 2024-10-24 17:20:12

[LintCode] Search a 2D Matrix的相关文章

LintCode &quot;Search a 2D Matrix II&quot;

Simply a revise of a genius Greedy algorithm seen on LeetCode - linear walking. class Solution { public: /** * @param matrix: A list of lists of integers * @param target: An integer you want to search in matrix * @return: An integer indicate the tota

[LeetCode] Search a 2D Matrix II 搜索一个二维矩阵之二

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.

leetcode Search a 2D Matrix II

题目连接 https://leetcode.com/problems/search-a-2d-matrix-ii/ Search a 2D Matrix II Description 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 ascend

【LeetCode】240. Search a 2D Matrix II

Search a 2D Matrix II 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 ascendin

leetcode——Search a 2D Matrix 二维有序数组查找(AC)

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 ro

[leedcode 240] Search a 2D Matrix II

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.

leetcode_74题——Search a 2D Matrix(数组查找)

Search a 2D Matrix Total Accepted: 40009 Total Submissions: 127082My Submissions Question Solution 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 fr

LeetCode: Search a 2D Matrix 解题报告

Search a 2D Matrix 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

[leetcode]Search a 2D Matrix @ Python

原题地址:https://oj.leetcode.com/problems/search-a-2d-matrix/ 题意: 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 o