[LeetCode]Search a 2D Matrix

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.

 

二分查找，时间复杂度O(logm)+O(logn)。

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

 

这题也可以从右上角遍历，时间复杂度O(m)+O(n)。

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