Leetcode 74. Search a 2D Matrix & 240. Search a 2D Matrix II

74. Search a 2D Matrix

Total Accepted: 92869 Total Submissions: 265460 Difficulty: Medium

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.

For example,

Consider the following matrix:

[  [1,   3,  5,  7],  [10, 11, 16, 20],  [23, 30, 34, 50]]

思路：

找行的时候是用的for loop，确定行了之后调用二分法查找。（解法1）

网上也有说法是把这个mn矩阵看成一行，那么共有mn个数，假设是第x个数 ( 0~mn-1)，那么对应的row: x/4, 对应的列数：x%4。 （解法2）

或者直接bs找行，然后再bs本行。（解法3）

public class Solution {    public boolean searchMatrix(int[][] m, int target){        if (m == null || m.length == 0) return false;                int row = m.length-1;        int col = m[0].length-1;                for (int i = 0 ; i <= row; i++){            if (m[i][0] <= target && m[i][col] >= target) return helper(m, 0, col, i, target);        }        return false;    }        boolean helper(int[][] m, int start, int end, int row, int target){        if (start > end) return false;                int mid = (start + end)/2;        if (m[row][mid] == target){            return true;        }else if (m[row][mid] > target){            return helper(m, 0, mid-1, row, target);        }else{            return helper(m, mid+1, end, row, target);        }    }}

240. Search a 2D Matrix II

Total Accepted: 48936 Total Submissions: 134289 Difficulty: Medium

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.

解法一：

二分法查找。与1不同的是，每当拿一个位置与之比之后，只能确定删除一个block，还剩下的区域至少要进行两次相同的查找才会出结果。

比如例子中，左上角是(0, 0) 右下角(4, 4)。那么第一次取到(2, 2) (9)。比如要找7， 7比9小，那么我们确定的是比9大的区域（以9为左上顶点的block）中没有我们要的元素；如果大于则相反，以该点为右下顶点的block没有目标元素。

剩下的区域可以取9的上边block：1 到 19；再取 3到21的block。 也可以先取得1到21，再取7到19。两次调用的结果要用或关系。

public class Solution { // 17ms    public boolean searchMatrix(int[][] matrix, int target) {        if(matrix == null || matrix.length ==0 || matrix[0].length == 0) return false;        return helper(matrix, target, 0, 0, matrix.length-1, matrix[0].length-1);    }        public boolean helper(int[][] matrix, int target, int x1, int y1, int x2, int y2){        if(x1 > x2 || y1 > y2) return false;        int mx = (x1 + x2)/2;        int my = (y1 + y2)/2;        if(matrix[mx][my] == target) return true;        else if(matrix[mx][my] > target) return helper(matrix, target, x1, y1, mx-1, y2) || helper(matrix, target, mx, y1, x2, my-1);        else return helper(matrix, target, mx+1, y1, x2, y2) || helper(matrix, target, x1, my+1, mx, y2);    }}

解法二：

来源这里。

从右上角开始, 比较target 和 matrix[i][j]的值. 如果小于target, 则该行不可能有此数,  所以i++; 如果大于target, 则该列不可能有此数, 所以j--. 遇到边界则表明该矩阵不含target.

真是神算法。。

public class Solution {  // 13ms    public boolean searchMatrix(int[][] matrix, int target) {          if(matrix.length == 0 || matrix[0].length == 0) return false;                    int i = 0, j = matrix[0].length - 1;                    while(i < matrix.length && j >= 0) {              int x = matrix[i][j];              if(target == x) return true;              else if(target < x) j--;              else i++;          }          return false;      }  }