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  • [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)。

     1 class Solution {
     2 public:
     3     bool searchMatrix(vector<vector<int>>& matrix, int target) {
     4         if(matrix.size()==0 || matrix[0].size()==0) return false;
     5         int m=matrix.size(),n=matrix[0].size();
     6         int left=0,right=m-1;
     7         while(left<=right)
     8         {
     9             int mid = (left+right)/2;
    10             if(matrix[mid][0]==target) return true;
    11             else if(matrix[mid][0]<target)
    12             {
    13                 left = mid+1;
    14             }
    15             else
    16             {
    17                 right = mid-1;
    18             }
    19         }
    20         int pos = right;
    21         if(pos<0) return false;
    22         left=0;
    23         right = n-1;
    24         while(left<=right)
    25         {
    26             int mid = (left+right)/2;
    27             if(matrix[pos][mid]==target) return true;
    28             else if(matrix[pos][mid]<target)
    29             {
    30                 left = mid+1;
    31             }
    32             else
    33             {
    34                 right = mid-1;
    35             }
    36         }
    37         return false;
    38     }
    39 };

    这题也可以从右上角遍历,时间复杂度O(m)+O(n)。

     1 class Solution {
     2 public:
     3     bool searchMatrix(vector<vector<int>>& matrix, int target) {
     4         if(matrix.size()==0 || matrix[0].size()==0) return false;
     5         int m=matrix.size(),n=matrix[0].size();
     6         int row=0,col=n-1;
     7         while(row<m && col>=0)
     8         {
     9             if(matrix[row][col]==target) return true;
    10             else if(matrix[row][col]<target) row++;
    11             else col--;
    12         }
    13         return false;
    14     }
    15 };
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  • 原文地址:https://www.cnblogs.com/Sean-le/p/4805532.html
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