[Leetcode]search a 2D Matrix II

Write an efficient algorithm, searches for a value in a m x n Matrix. This matrix has the following properties:

• Integers in each row is sorted in ascending from left to right.
• Integers in each column is sorted in ascending from top to bottom.

For example,

Consider the following matrix:

[[1, 4, 7, 11, 15], [2, 5, 8, 12, 19], [3, 6, 9, 16, 22], [10, 13, 14, 17 , [+], [approx., +, +]] 

given target  = 5 , Return true .

Given Target = 20 , return false .

[Idea]o (m+n) complexity, starting from the top of the search. Because the matrix is ordered, you can narrow the scope based on the size relationship.

Class Solution {public:    bool Searchmatrix (vector<vector<int>>& matrix, int target) {        int m = Matrix.size ();        int n = matrix[0].size ();        int i = 0;        int j = n-1;        while (i>=0 && j>=0 && i<m &&j<n) {            if (target = = Matrix[i][j])                return true;< C8/>if (Target>matrix[i][j]) {                i++;            }            else                j--;        }        return false;    }};

O (MLOGN) searches each row for two points.

Class Solution {public:    bool Searchmatrix (vector<vector<int>>& matrix, int target) {        int m = Matrix.size ();        int n = matrix[0].size ();        for (int i=0; i<m; ++i) {            if (matrix[i][0]<=target && matrix[i][n-1]>=target) {                if ( Searchvector (matrix[i],target))                    return true;            }        }        return false;    }       BOOL Searchvector (vector<int>& v, int target) {    int left = 0, right = v.size ()-1;    while (left <= right) {        int mid = left + (right-left)/2;        if (v[mid] = = target)            return true;        if (V[mid] < target) Left            = mid + 1;        else right         = mid-1;        }        return false;    }    };

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[Leetcode]search a 2D Matrix II