[leetcode] Search a 2D Matrix

最新推荐文章于 2020-05-02 19:57:48 发布

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本文介绍了一种高效的算法，用于在一个m x n的矩阵中查找特定值。该矩阵每一行的整数从左到右递增排序，并且每行的第一个整数大于前一行的最后一个整数。文章提供了一个具体的示例矩阵及目标值，并展示了如何使用C++实现搜索过程。

From : https://leetcode.com/problems/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.

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        if(!matrix.size() || !matrix[0].size()) return false;
        int startRow=0, start=0, endRow=matrix.size()-1, end=matrix[0].size()-1, midRow = 0;
        while(startRow <= endRow) {
            midRow = (startRow+endRow)>>1;
			if(matrix[midRow][start] <= target && matrix[midRow][end] >= target) {
				break;
			}
            else if(matrix[midRow][end] < target) {
                startRow = midRow+1;                
            } else {
                endRow = midRow-1;
            }
        }
		if(matrix[midRow][start] > target || matrix[midRow][end] < target) return false;
        while(start <= end) {
            int mid = (start+end)>>1;
            if(matrix[midRow][mid] == target) return true;
            else if(matrix[midRow][mid] > target) {
                end = mid-1;
            } else {
                start = mid+1;
            }
        }
        return false;
    }
};