[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<1) return false;

        int n=matrix[0].size();

        int tm;

//cout<<"m:"<<m<<" n:"<<n<<endl;

        if(m==1)    tm=0;

        else{

            if(matrix[m-1][0]<=target)   tm=m-1;

            else{

                int lft=0,rgt=m-1;

                tm =0;

                do{

                    if(matrix[tm+1][0]>target)  break;

                    int mid = (lft+rgt)/2;

                    if(matrix[mid][0]>target)   rgt=mid;

                    else    lft=mid;

                    tm = lft;

                }while(lft+1<rgt);

            }

        }

        int lft=0,rgt=n-1;

        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)/2;

            if(matrix[tm][mid]>target)  rgt = mid;

            else lft=mid;

            tn = lft;

            if(matrix[tm][tn]==target)  return true;

        }while(lft+1<rgt);

        return false;

    }

};



int main()

{

    vector<vector<int> > matrix{{1,   3,  5,  7},{10, 11, 16, 20},{23, 30, 34, 50}};

    Solution sol;

    cout<<sol.searchMatrix(matrix,2)<<endl;

    return 0;

}

View Code