LintCode: Search A 2d Matrix

1.

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）

 class Solution {
 public:
     /**
      * @param matrix, a list of lists of integers
      * @param target, an integer
      * @return a boolean, indicate whether matrix contains target
      */
     bool searchMatrix(vector<vector<int> > &matrix, int target) {
         // write your code here
         int m = matrix.size();
         if (m == ) {
             return false;
         }
         int n = matrix[].size();
         int r = , c = n - ;
         while (r < m && c >= ) {
             if (matrix[r][c] < target) {
                 r++;
             } else if (matrix[r][c] > target) {
                 c--;
             } else {
                 return true;
             }
         }
         return false;
     }
 };

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

 class Solution {
 public:
     /**
      * @param matrix, a list of lists of integers
      * @param target, an integer
      * @return a boolean, indicate whether matrix contains target
      */
     bool helper(vector<vector<int> > &M, int x1, int y1, int x2, int y2, int target) {
         if (x1 > x2 || y1 > y2) {//empty matrix
             return false;
         }
         int midx = (x1 + x2)>>;
         int midy = (y1 + y2)>>;
         if (M[midx][midy] == target) {
             return true;
         }
         return (M[midx][midy] > target)?
         (helper(M, x1, y1, x2, midy-, target)||helper(M, x1, midy, midx-, y2, target)):
         (helper(M, x1, midy+, x2, y2, target)||helper(M, midx+, y1, x2, midy, target));

     }
     bool searchMatrix(vector<vector<int> > &matrix, int target) {
         // write your code here
         int m = matrix.size();
         if (m == ) {
             return false;
         }
         int n = matrix[].size();
         return helper(matrix, , , m - , n - , target);
     }
 };

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)

 class Solution {
 public:
     /**
      * @param A, a list of integers
      * @param left, an integer
      * @param right, an integer
      * @param target, an integer
      * @return an integer, indicate the index of the last number less than or equal to target in A
      */
     int binarySearch(vector<int> &A, int left, int right, int target) {
         while (left <= right) {//not an empty list
             int mid = (left + right) >> ;
             if (A[mid] <= target) {
                 left = mid + ;//left is in the integer after the last integer less than or equal to target
             } else {
                 right = mid - ;
             }
         }
         return left - ;
     }
     /**
      * @param M, a list of lists of integers
      * @param x1, an integer
      * @param y1, an integer
      * @param x2, an integer
      * @param y2, an integer
      * @param target, an integer
      * @return a boolean, indicate whether matrix contains target
      */
     bool helper(vector<vector<int> > &M, int x1, int y1, int x2, int y2, int target) {
         if (x1 > x2 || y1 > y2) {//empty matrix
             return false;
         }
         int midx = (x1 + x2)>>;
         int indy = binarySearch(M[midx], y1, y2, target);
         //M[midx][indy] <= target
         if ((indy >= y1) && (M[midx][indy] == target)) {
             return true;
         }
         return (helper(M, x1, indy+, midx-, y2, target))||(helper(M, midx+, y1, x2, indy, target));

     }
     /**
      * @param matrix, a list of lists of integers
      * @param target, an integer
      * @return a boolean, indicate whether matrix contains target
      */
     bool searchMatrix(vector<vector<int> > &matrix, int target) {
         // write your code here
         int m = matrix.size();
         if (m == ) {
             return false;
         }
         int n = matrix[].size();
         return helper(matrix, , , m - , n - , target);
     }
 };