[LeetCode]Search a 2D Matrix

最新推荐文章于 2020-05-02 19:57:48 发布

原创最新推荐文章于 2020-05-02 19:57:48 发布 · 932 阅读

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本文介绍了一种高效的搜索算法，用于在一个有序的二维矩阵中查找特定的目标值。该算法首先通过二分查找确定目标可能存在的行，然后在该行内再次使用二分查找找到目标值。这种方法的时间复杂度为O(log n)，显著提高了搜索效率。

class Solution {
//binary search at the last column first, find the interested row
//then binary search in this row
public:
	bool searchMatrix(vector<vector<int> > &matrix, int target) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		int rowNum = matrix.size();
		if(0 == rowNum) return false;
		int colNum = matrix[0].size();
		//get the row 
		int l = 0; int r = rowNum-1;
		while (l <= r)//find the first first element bigger than target
		{
			int m = l+(r-l)/2;
			if (matrix[m][colNum-1] == target)
				return true;
			else if (matrix[m][colNum-1] < target)
				l = m+1;
			else// if ((matrix[m][colNum-1] > target)
				r = m-1;
		}
		int row = l;
		if(row < 0 || row >= rowNum) return false;
		l = 0; r = colNum-1;
		while (l <= r)
		{
			int m = l+(r-l)/2;
			if (matrix[row][m] == target)
				return true;
			else if (matrix[row][m] < target)
				l = m+1;
			else// if ((matrix[m][colNum-1] > target)
				r = m-1;
		}
		return false;
	}
};

second time

class Solution {
public:
    int searchRow(vector<vector<int> > &matrix, int target)
    {
        int rowNum = matrix.size();
        int colNum = matrix[0].size();
        int left = 0;
        int right = rowNum-1;
        while(left <= right)
        {
            int mid = left+(right-left)/2;
            int curNum = matrix[mid][colNum-1];
            if(curNum == target) return mid;
            else if(curNum > target) right = mid-1;
            else left = mid+1;
        }
        if(left >= 0 && left < rowNum && matrix[left][colNum-1] >= target) return left;
        else return false;
    }
    int searchColumn(vector<vector<int> > &matrix, int rowIdx, int target)
    {
        int rowNum = matrix.size();
        int colNum = matrix[0].size();
        int left = 0;
        int right = colNum-1;
        while(left <= right)
        {
            int mid = left+(right-left)/2;
            int curNum = matrix[rowIdx][mid];
            if(curNum == target) return mid;
            else if(curNum > target) right = mid-1;
            else left = mid+1;
        }
        return -1;
    }
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int n = matrix.size();
        if(n == 0) return false;
        int m = matrix[0].size();
        if(m == 0) return false;
        
        int rowIdx = searchRow(matrix, target);
        if(rowIdx == -1) return false;
        int columnIdx = searchColumn(matrix, rowIdx, target);
        if(columnIdx == -1) return false;
        return true;
    }
};