Search a 2D Matrix

第一次

两次二分。思路比较简单，但是要注意边界条件。

 bool searchMatrix(vector<vector<int> > &matrix, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int m = matrix.size();
        if(m == 0)
            return false;
        int n = matrix[0].size();
        if(n == 0)
            return false;
        int l = 0, r = m, mid, tmp;
        while(l != r){
            mid = (l+r)/2;
            if(target == matrix[mid][0])
                return true;
            else if(target < matrix[mid][0]){
                r = mid;
            }
            else{
                l = mid+1;
            }
        }
        if(l > m-1)
            l = m-1;
        if(target < matrix[l][0])
            tmp = l>1?l-1:0;
        else
            tmp = l<m-2?l+1:m-1;
        l = 0;
        r = n;
        while(l != r){
            mid = (l+r)/2;
            if(target == matrix[tmp][mid])
                return true;
            else if(target < matrix[tmp][mid])
                r = mid;
            else
                l = mid+1;
        }
        return false;
    }

 第二次

每次判断左下角的数，然后删除一行或一列，复杂度O（m+n）

 bool searchMatrix(vector<vector<int> > &matrix, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int m = matrix.size();
        if(m == 0)
            return false;
        int n = matrix[0].size();
        if(n == 0)
            return false;
        int i = m-1, j = 0;
        while(i >= 0 && j < n){
            if(matrix[i][j] == target)
                return true;
            else if(matrix[i][j] < target)
                j++;
            else
                i--;
        }
        return false;
    }