[LeetCode] Search a 2D Matrix 二分搜索

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.

Hide Tags

Array Binary Search

 

---

　　　在一个矩阵中确定是否存在某个数，这个矩阵有规律：一个排序好的数组，一行一行地填入矩阵。解题思路就是先二分查找会在哪一行，然后二分查找找是否存在。

#include<iostream>
#include<vector>
using namespace std;

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        int m=matrix.size();
        if(m<) return false;
        int n=matrix[].size();
        int tm;
//cout<<"m:"<<m<<" n:"<<n<<endl;
        if(m==)    tm=;
        else{
            if(matrix[m-][]<=target)   tm=m-;
            else{
                int lft=,rgt=m-;
                tm =;
                do{
                    if(matrix[tm+][]>target)  break;
                    int mid = (lft+rgt)/;
                    if(matrix[mid][]>target)   rgt=mid;
                    else    lft=mid;
                    tm = lft;
                }while(lft+<rgt);
            }
        }
        int lft=,rgt=n-;
        if(matrix[tm][rgt]==target||matrix[tm][lft]==target) return  true;
        if(matrix[tm][rgt]<target)  return false;
        int tn=lft;
        do{
            int mid=(lft+rgt)/;
            if(matrix[tm][mid]>target)  rgt = mid;
            else lft=mid;
            tn = lft;
            if(matrix[tm][tn]==target)  return true;
        }while(lft+<rgt);
        return false;
    }
};

int main()
{
    vector<vector<int> > matrix{{,   ,  ,  },{, , , },{, , , }};
    Solution sol;
    cout<<sol.searchMatrix(matrix,)<<endl;
    return ;
}