240. Search a 2D Matrix II（二分查找）

 

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 in ascending from left to right.
• Integers in each column are sorted in ascending from top to bottom.

Example:

Consider the following matrix:

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

Given target = 5, return true.

Given target = 20, return false.

 

 

 

class Solution:
    def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:
        i = len(matrix) - 1 
        j = 0
        while j < len(matrix[0]) and i >=0:
            if matrix[i][j] > target:
                i-=1
            elif matrix[i][j] < target:
                j+=1
            else:
                return True
        return False

 

 

 

class Solution:
    def searchInsert(self, nums: List[int], target: int) -> int:
        lo = 0
        hi = len(nums) - 1
        while lo <=hi:
            mid = lo + (hi-lo)//2
            if target < nums[mid]:
                hi = mid - 1
            elif target > nums[mid]:
                lo = mid + 1
            else:
                return mid
        return lo

 

3、可以每行都用二分查找。

class Solution {
public:
     bool searchInsert(vector<int>& nums, int target) {
        int low = 0, high = nums.size()-1;
        while(low <= high) {
            int mid = low + (high - low)/2;
            if ( target > nums[mid]) {
                low = mid +1;
            } else if (target < nums[mid]){
                high = mid -1;
            }  else {
                return true ;
            }
        }
        return false;
    }
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        for (auto& vec : matrix) {
            if (searchInsert(vec,target)) {
                return true;
            }
        }
        return false;
    }
};