[LeetCode] 74. Search a 2D Matrix 搜索一个二维矩阵

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.

在一个给定的矩阵里查找某一个值，矩阵有如下特点：每一行里的整数都是按从左到右排序；每一行的第一数都大于前一行的最后一个数。

解法：二分法Binary Search，利用给定矩阵的特点，Z形大小顺序排列，把矩阵变成数组，使用二分法查找，或者直接在矩阵上使用二分法。

Python：

class Solution(object):
    def searchMatrix(self, matrix, target):
        """
        :type matrix: List[List[int]]
        :type target: int
        :rtype: bool
        """
        if not matrix:
            return False
        
        m, n = len(matrix), len(matrix[0])
        left, right = 0, m * n
        while left < right:
            mid = left + (right - left) / 2
            if matrix[mid / n][mid % n] >= target:
                right = mid
            else:
                left = mid + 1

        return left < m * n and matrix[left / n][left % n] == target

C++:  Time: O(logm + logn)， Space: O(1)

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        if (matrix.empty()) {
            return false;
        }

        // Treat matrix as 1D array.
        const int m = matrix.size();
        const int n = matrix[0].size();
        int left = 0;
        int right = m * n - 1;

        // Find min of left s.t.  matrix[left / n][left % n] >= target
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (matrix[mid / n][mid % n] >= target) {
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        }

        // Check if matrix[left / n][left % n] equals to target.
        if (left != m * n && matrix[left / n][left % n] == target) {
            return true;
        }

        return false;
    }
};　　

C++:

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        if (matrix.empty() || matrix[0].empty()) return false;
        if (target < matrix[0][0] || target > matrix.back().back()) return false;
        int m = matrix.size(), n = matrix[0].size();
        int left = 0, right = m * n - 1;
        while (left <= right) {
            int mid = (left + right) / 2;
            if (matrix[mid / n][mid % n] == target) return true;
            else if (matrix[mid / n][mid % n] < target) left = mid + 1;
            else right = mid - 1;
        }
        return false;
    }
};

　　

　　

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