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; }};
