[leetcode] Search a 2D Matrix

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]
]

Given target = 3, return true.

https://oj.leetcode.com/problems/search-a-2d-matrix/

 

二分法。

1. 先二分在第一列（或者第一行）找出元素可能在的行，然后二分查找该列。

2. 二维数组转为一维数组，然后利用 /，%求坐标二分查找。

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0)
            return false;
        int m = matrix.length;
        int n = matrix[0].length;
        int left = 0, right = m - 1;
        int mid;
        while (left <= right) {
            mid = (left + right) / 2;
            if (target > matrix[mid][0]) {
                left = mid + 1;
            } else if (target < matrix[mid][0])
                right = mid - 1;
            else {
                return true;
            }
        }
        int row = right;
        if (row < 0 || row >= m)
            return false;

        // not found, keep searching
        left = 0;
        right = n - 1;
        while (left <= right) {
            mid = (left + right) / 2;
            if (target > matrix[row][mid])
                left = mid + 1;
            else if (target < matrix[row][mid])
                right = mid - 1;
            else
                return true;
        }

        return false;

    }

    public boolean searchMatrix2(int[][] matrix, int target) {

        if (matrix.length == 0 || matrix[0].length == 0) {
            return false;
        }

        int xLength = matrix[0].length;
        int min = 0;
        int max = matrix[0].length * matrix.length - 1;

        int x, y, cur;
        while (min <= max) {
            cur = (min + max) / 2;
            y = cur / xLength;
            x = cur % xLength;
            if (matrix[y][x] == target) {
                return true;
            } else if (target < matrix[y][x]) {
                max = cur - 1;
            } else {
                min = cur + 1;
            }
        }

        return false;
    }

    public static void main(String[] args) {
        int[][] matrix = new int[][] { { 1, 3, 5, 7 }, { 10, 11, 16, 20 }, { 23, 30, 34, 50 } };
        System.out.println(new Solution().searchMatrix(matrix, 7));
        System.out.println(new Solution().searchMatrix2(matrix, 7));

    }

}

 

 

扩展：如果说不保证每行第一个数据比上一行最后一个大，只满足比同列上面行的数据大，该如何处理。

即

　　Table[i][j] ≤ Table[i][j + 1],
　　Table[i][j] ≤ Table[i + 1][j]

则，此时上述2分法不起作用了，leetcode论坛里这个题目的讨论讲的很详细，列举了很多错误的解法和正确的解法，看这里。

随便举个反例，下面的矩阵找9或者找10。

{ 1, 4, 7, 9, 15 },

{ 2, 5, 8, 12, 19 },

{ 3, 6, 10, 16, 22 },

{ 10, 13, 14, 17, 24 },

{ 18, 21, 23, 26, 30 }

 

里面最好的两种解法。假设m==n方便讨论。

思路1：从右上角（或者左下角）开始，行列两个指针，如果当前元素比target小，向下移动行指针；如果当前元素比target大，向左移动列指针，直到找到元素或者失败。最坏情况移动对角线长度，复杂度O(n)。

思路2：优化的二分查找。每次在中间行（或者列）二分查到找所在范围，如果没查到继续查找左下和右上两个小矩阵（左上和右下不可能存在）。 T(n)=2T(n/2)+clogn。根据主定律可求出复杂度也是O(n)。

 

思路1代码：

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

 

思路2代码：（有空写）

 

 

第二遍记录：转化成一维的二分查找

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix==null||matrix.length==0||matrix[0].length==0)
            return false;
        int m = matrix.length;
        int n = matrix[0].length;
        int left=0;
        int right=m*n-1;
        while(left<=right){
           int mid=left+(right-left)/2;
           int x =mid/n;
           int y =mid%n;
           if(target>matrix[x][y]){
               left=mid+1;
           }else if(target<matrix[x][y]){
               right=mid-1;
           }
           else
                return true;
            
        }
        return false;
    }
}