【LeetCode】74. Search a 2D Matrix 解題報告

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74. Search a 2D Matrix / Medium

Write an efficient algorithm that searches for a value target in an m x n integer matrix 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.

Example 1:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
Output: true

Example 2:

Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
Output: false

Constraints:

  • m == matrix.length
  • n == matrix[i].length
  • 1 <= m, n <= 100
  • -10^4 <= matrix[i][j], target <= 10^4

Solution: Binary Search

思路

這題不難，其實就是把整個二維陣列當成一維陣列，再用 Binary Search 做搜尋。
至於如何計算出 Index，row 的部分是直接除，col 的部分則是餘數。

效能

Complexity

  • Time Complexity: O(log(mn))
  • Space Complexity: O(1)

LeetCode Result

Code

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