LeetCode #240: Search a 2D Matrix II

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 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.

Given target = 20, return false.

Solution

题意

给定一个二维矩阵，该矩阵的行和列都是各自递增的。实现一个查找算法。

分析

由于本题的分类为Divide and Conquer，所以先考虑用递归的解法（但是不用递归代码反而更简洁）。

递归解法：

跟递归有关的首先想到的是二分查找，分析矩阵的性质可知，对于处在矩阵中心的数字x，如果：
1. target == x，查找成功，返回true
2. target > x，则以x为端点的左上角的子矩阵中一定不含target（左上角的子矩阵的值都小于x）
3. target < x，则以x为端点的右下角的子矩阵中一定不含target（右下角的子矩阵的值都大于x）
这样每递归一次就可以把查找范围缩小1/4，代码中要注意边界条件。

非递归解法：

从矩阵的左上角或者右下角开始查找，以左上角为例（此时i = 0, j = matrix[0].size() - 1），对于x = matrix[i][j]：
1. target == x，查找成功，返回true
2. target > x，则该行一定不包含target，i++
3. target < x，则该列一定不包含target，j–

Code

Version 1 - 递归版

class Solution {public:    bool searchMatrix(vector<vector<int>>& matrix, int target) {        if (matrix.empty() || matrix[0].empty())            return false;        int rsize = matrix.size();        int csize = matrix[0].size();        return aux(matrix, target, 0, rsize - 1, 0, csize - 1);    }    bool aux(vector<vector<int>>& matrix, int target, int rlow, int rhigh, int clow, int chigh) {        if (rlow > rhigh || clow > chigh)            return false;        if (rlow == rhigh && clow == chigh)            return matrix[rlow][clow] == target;        int rmid = (rlow + rhigh) / 2;        int cmid = (clow + chigh) / 2;        if (matrix[rmid][cmid] == target)            return true;        else if (matrix[rmid][cmid] > target)            return aux(matrix, target, rlow, rmid - 1, clow, chigh) ||                    aux(matrix, target, rmid, rhigh, clow, cmid - 1);        else            return aux(matrix, target, rmid + 1, rhigh, clow, chigh) ||                    aux(matrix, target, rlow, rmid, cmid + 1, chigh);    }};

Version 2 - 非递归版

class Solution {public:    bool searchMatrix(vector<vector<int>>& matrix, int target) {        if (matrix.empty() || matrix[0].empty())            return false;        int i_max = matrix.size();        int i = 0;        int j = matrix[0].size() - 1;        while (i < i_max && j >= 0) {            int x = matrix[i][j];            if (target == x)                return true;            else if (target > x)                i++;            else                j--;        }        return false;    }};