Leetcode_c++： Search a 2D Matrix (074）

题目

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.

---

算法

复杂度：O（lg（nm））
二分法确定target可能在第几行出现。再用二分法在该行确定target可能出现的位置。时间复杂度O(logn+logm)

---

class Solution {  
public:  
    bool searchMatrix(vector<vector<int> > &matrix, int target) {  
        int left = 0;  
        int right = matrix.size()-1;  
        if(left != right)  
        {  
            while(left<=right)  
            {  
                int middle = left + (right-left)/2;  
                if(matrix[middle][0] < target)  
                {  
                    left = middle+1;  
                }  
                else if(matrix[middle][0] > target)  
                {  
                    right = middle-1;  
                }  
                else  
                {  
                    return true;  
                }  
            }  
        }  
        if(right == -1)  
        {  
            return false;  
        }  
        else  
        {  
            int row = right;  
            int left = 0;  
            int right = matrix[row].size()-1;  
            while(left<=right)  
            {  
                int middle = left + (right-left)/2;  
                if(matrix[row][middle] < target)  
                {  
                    left = middle+1;  
                }  
                else if(matrix[row][middle] > target)  
                {  
                    right = middle-1;  
                }  
                else  
                {  
                    return true;  
                }  
            }  
            return false;  
        }  

    }  
};  

//二分查找的简单版本
class Solution {
public:
  bool searchMatrix(const vector<vector<int> > &matrix, int target){
    if (matrix.empty()) return false;
    int m = matrix.size(), n = matrix.front().size();
    int begin = 0, end = m * n, middle, row, col;


    while(begin < end){
      middle = begin + (end - begin) / 2;
      int row = middle / n;
      int col = middle % n;
      if(matrix[row][col] == target) return true;
      else if(matrix[row][col] < target) begin = middle + 1;
      else end = middle;
    }
    return false;
  }
};

---

算法

从右上角元素开始遍历，每次遍历中若与target相等则返回true；若小于则行向下移动；若大于则列向左移动。时间复杂度m+n

---

class Solution {  
public:  
    bool searchMatrix(vector<vector<int> > &matrix, int target) {  
        int i =matrix.size()-1; int j=0;  
        int n = matrix.size(); int m = matrix[0].size();  
        while(i>=0&&jif(matrix[i][j]==target)  
            {  
                return true;  
            }  
            else if(matrix[i][j]else  
            {  
                i--;  
            }  
        }  
        return false;  
    }