LintCode: Search A 2d Matrix

设查找的数位y，第一行最后一列的数位x

如果x<y，x是第一行最大的，所以第一行都小于y，删除第一行；

如果x>y，x是最后一列最小的，所以最后一列都大于y，删除最后一列；

这样保证x永远在可能有解的矩阵的第一行，最后一列。

时间复杂度：O（m＋n）

 1 class Solution {
 2 public:
 3     /**
 4      * @param matrix, a list of lists of integers
 5      * @param target, an integer
 6      * @return a boolean, indicate whether matrix contains target
 7      */
 8     bool searchMatrix(vector<vector<int> > &matrix, int target) {
 9         // write your code here
10         int m = matrix.size();
11         if (m == 0) {
12             return false;
13         }
14         int n = matrix[0].size();
15         int r = 0, c = n - 1;
16         while (r < m && c >= 0) {
17             if (matrix[r][c] < target) {
18                 r++;
19             } else if (matrix[r][c] > target) {
20                 c--;
21             } else {
22                 return true;
23             }
24         }
25         return false;
26     }
27 };

2. 分治法1

设查找的数位y，取中心点x，把矩阵分解成4部分

如果x<y，x是A中最大的，所以A都小于y，删除A；

如果x>y，x是D中最小的，所以D都小于y，删除D；

A ｜ B

————

C ｜ D

时间复杂度：O(N) = O(N/4)+O(N/2)+O(1), 介于O(N^0.5)~O(N)之间

 1 class Solution {
 2 public:
 3     /**
 4      * @param matrix, a list of lists of integers
 5      * @param target, an integer
 6      * @return a boolean, indicate whether matrix contains target
 7      */
 8     bool helper(vector<vector<int> > &M, int x1, int y1, int x2, int y2, int target) {
 9         if (x1 > x2 || y1 > y2) {//empty matrix
10             return false;
11         }
12         int midx = (x1 + x2)>>1;
13         int midy = (y1 + y2)>>1;
14         if (M[midx][midy] == target) {
15             return true;
16         }
17         return (M[midx][midy] > target)?
18         (helper(M, x1, y1, x2, midy-1, target)||helper(M, x1, midy, midx-1, y2, target)):
19         (helper(M, x1, midy+1, x2, y2, target)||helper(M, midx+1, y1, x2, midy, target));
20         
21     }
22     bool searchMatrix(vector<vector<int> > &matrix, int target) {
23         // write your code here
24         int m = matrix.size();
25         if (m == 0) {
26             return false;
27         }
28         int n = matrix[0].size();
29         return helper(matrix, 0, 0, m - 1, n - 1, target);
30     }
31 };

3. 分治法2

设查找的数为y，在中线找到这样两个数x1，x2，使得x1<y,x2>y，把矩阵分成4部分

A｜ B

————

C｜ D

x1是A中最大的，所以A都小于y，删掉A；

x2是D中最小的，所以D都大于y，删掉D；

时间复杂度：O(N)＝2O(N/4)+O(logn), 为O(N^0.5)

 1 class Solution {
 2 public:
 3     /**
 4      * @param A, a list of integers
 5      * @param left, an integer
 6      * @param right, an integer
 7      * @param target, an integer
 8      * @return an integer, indicate the index of the last number less than or equal to target in A
 9      */
10     int binarySearch(vector<int> &A, int left, int right, int target) {
11         while (left <= right) {//not an empty list
12             int mid = (left + right) >> 1;
13             if (A[mid] <= target) {
14                 left = mid + 1;//left is in the integer after the last integer less than or equal to target 
15             } else {
16                 right = mid - 1;
17             }
18         }
19         return left - 1;
20     }
21     /**
22      * @param M, a list of lists of integers
23      * @param x1, an integer
24      * @param y1, an integer
25      * @param x2, an integer
26      * @param y2, an integer
27      * @param target, an integer
28      * @return a boolean, indicate whether matrix contains target
29      */
30     bool helper(vector<vector<int> > &M, int x1, int y1, int x2, int y2, int target) {
31         if (x1 > x2 || y1 > y2) {//empty matrix
32             return false;
33         }
34         int midx = (x1 + x2)>>1;
35         int indy = binarySearch(M[midx], y1, y2, target);
36         //M[midx][indy] <= target
37         if ((indy >= y1) && (M[midx][indy] == target)) {
38             return true;
39         }
40         return (helper(M, x1, indy+1, midx-1, y2, target))||(helper(M, midx+1, y1, x2, indy, target));
41         
42     }
43     /**
44      * @param matrix, a list of lists of integers
45      * @param target, an integer
46      * @return a boolean, indicate whether matrix contains target
47      */
48     bool searchMatrix(vector<vector<int> > &matrix, int target) {
49         // write your code here
50         int m = matrix.size();
51         if (m == 0) {
52             return false;
53         }
54         int n = matrix[0].size();
55         return helper(matrix, 0, 0, m - 1, n - 1, target);
56     }
57 };