[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.

 

从右上角向左下角搜索，初始row = 0, col = n - 1; 如果当前值等于目标值，返回真，如果以目标值大，则舍弃当前列，否则舍弃当前行。

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        int row = 0, col = matrix[0].size() - 1;
        while (row < matrix.size() && col >=0) {
            if (matrix[row][col] == target) {
                return true;
            } else if (matrix[row][col] > target) {
                --col;
            } else {
                ++row;
            }
        }
        return false;
    }
};

 第二遍做的时候发现其实这个矩阵不完全是杨氏矩阵，可以使用二分搜索，下面是代码。

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        int m = matrix.size();
        int n = matrix[0].size();
        int L = 0, R = n * m - 1, M;
        int col, row;
        while (L <= R) {
            M = L + ((R - L) >> 1);
            row = M / n;
            col = M % n;
            if (matrix[row][col] > target) R = M - 1;
            else if (matrix[row][col] < target) L = M + 1;
            else return true;
        }
        return false;
    }
};

 上面是对所有元素二分，复杂度是log(n*m)，还可以更优化一点，对于普通的杨氏矩阵也适用，复杂度为log(n)+log(m).

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        int row = lowerBound(matrix, target);
        if (row == -1) return false;
        return binSearch(matrix, target, row);
    }
    
    int lowerBound(vector<vector<int> > &matrix, int target) {
        int L = 0, R = matrix.size() - 1, M;
        while (L <= R) {
            M = L + ((R - L) >> 1);
            if (matrix[M][0] <= target) L = M + 1;
            else R = M - 1;
        }
        return R;
    }
    
    bool binSearch(vector<vector<int> > &matrix, int target, int row) {
        int L = 0, R = matrix[row].size() - 1, M;
        while (L <= R) {
            M = L + ((R - L) >> 1);
            if (matrix[row][M] < target) L = M + 1;
            else if (matrix[row][M] > target) R = M - 1;
            else return true;
        }
        return false;
    }
};