算法 简单 | 28. 搜索二维矩阵

题目描述

写出一个高效的算法来搜索 m × n矩阵中的值。
这个矩阵具有以下特性：

1. 每行中的整数从左到右是排序的。
2. 每行的第一个数大于上一行的最后一个整数。

样例1

输入: [[5]],2
输出: false
样例解释: 没有包含，返回false。

样例2

输入:  
[
  [1, 3, 5, 7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
],3
输出: true
样例解释: 包含则返回true。

解题思路

可以看作是一个有序数组被分成了n段，每段就是一行。因此依然可以二分求解。
对每个数字，根据其下标i，j进行编号，每个数字可被编号为0～n*n-1

相当于是在一个数组中的下标。然后直接像在数组中二分一样来做。取的mid要还原成二位数组中的下标，i = mid/n, j = mid%n

java题解

// Binary Search Twice
public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0) {
            return false;
        }
        if (matrix[0] == null || matrix[0].length == 0) {
            return false;
        }
        
        int row = matrix.length;
        int column = matrix[0].length;
        
        // find the row index, the last number <= target 
        int start = 0, end = row - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (matrix[mid][0] == target) {
                return true;
            } else if (matrix[mid][0] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if (matrix[end][0] <= target) {
            row = end;
        } else if (matrix[start][0] <= target) {
            row = start;
        } else {
            return false;
        }
        
        // find the column index, the number equal to target
        start = 0;
        end = column - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (matrix[row][mid] == target) {
                return true;
            } else if (matrix[row][mid] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if (matrix[row][start] == target) {
            return true;
        } else if (matrix[row][end] == target) {
            return true;
        }
        return false;
    }
}

// Binary Search Once
public class Solution {
    /**
     * @param matrix, a list of lists of integers
     * @param target, an integer
     * @return a boolean, indicate whether matrix contains target
     */
    public boolean searchMatrix(int[][] matrix, int target) {
        // write your code here
        if(matrix == null || matrix.length == 0){
            return false;
        }
        
        if(matrix[0] == null || matrix[0].length == 0){
            return false;
        }
        
        int row = matrix.length;
        int column = matrix[0].length;
        
        int start = 0, end = row * column - 1;
        while(start <= end){
            int mid = start + (end - start) / 2;
            int number = matrix[mid / column][mid % column];
            if(number == target){
                return true;
            }else if(number > target){
                end = mid - 1;
            }else{
                start = mid + 1;
            }
        }
        
        return false;
        
    }
}

C++题解

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) {
        int n = matrix.size();
        if (n == 0) {
            return false;
        }
        
        int m = matrix[0].size();
        if (m == 0) {
            return false;
        }
        
        int start = 0, end = n * m - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            int row = mid / m;
            int col = mid % m;
            
            if (matrix[row][col] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        
        if (matrix[start / m][start % m] == target) {
            return true;
        }
        
        if (matrix[end / m][end % m] == target) {
            return true;
        }
        
        return false;
    }
};

python题解

class Solution:
    """
    @param matrix, a list of lists of integers
    @param target, an integer
    @return a boolean, indicate whether matrix contains target
    """
    def searchMatrix(self, matrix, target):
        if len(matrix) == 0:
            return False
            
        n, m = len(matrix), len(matrix[0])
        start, end = 0, n * m - 1
        while start + 1 < end:
            mid = (start + end) / 2
            x, y = mid / m, mid % m
            if matrix[x][y] < target:
                start = mid
            else:
                end = mid
        x, y = start / m, start % m
        if matrix[x][y] == target:
            return True
        
        x, y = end / m, end % m
        if matrix[x][y] == target:
            return True
        
        return False