刷题240. Search a 2D Matrix II

一、题目说明

题目240. Search a 2D Matrix II，从一个m*n的二维矩阵查找一个整数，每一行从左到右递增，每一列从上到下递增。

二、我的解答

先计算矩阵中点matrix[row_mid][col_mid]，然后将矩阵分成4个区间：

class Solution{
	public:
		bool dfs(vector<vector<int> >& matrix,int target,int start_x, int end_x, int start_y, int end_y){
			if(start_x > end_x || start_y > end_y)
	            return false;
	        if(start_x == end_x && start_y == end_y)
	            return matrix[start_x][start_y] == target;
	        if(target < matrix[start_x][start_y] || target > matrix[end_x][end_y])
	            return false;
	        int mid_x = start_x + (end_x - start_x) / 2;
	        int mid_y = start_y + (end_y - start_y) / 2;
	        if(matrix[mid_x][mid_y] == target) {
	            return true;
	        } else if(matrix[mid_x][mid_y] > target) {
	        	//在1,2,3区域找 
	            return dfs(matrix, target, start_x, mid_x - 1, start_y, mid_y - 1)
	                || dfs(matrix, target, mid_x, end_x, start_y, mid_y - 1)
	                || dfs(matrix, target, start_x, mid_x - 1, mid_y, end_y);
	        } else {
	        	//在2,3,4区域找 
	            return dfs(matrix, target, mid_x + 1, end_x, mid_y + 1, end_y)
	                || dfs(matrix, target, start_x, mid_x, mid_y + 1, end_y)
	                || dfs(matrix, target, start_x + 1, end_x, start_y, mid_y);
	        }
    	}

		bool searchMatrix(vector<vector<int> >& matrix,int target){
			if(matrix.size()<1){
				return false;
			}
			row = matrix.size();
			col = matrix[0].size();
			bool result = dfs(matrix,target,0,row-1,0,col-1);
			if(result) return true;
			else return false;
		}
	private:
		int row,col;
};

性能如下：

Runtime: 148 ms, faster than 24.09% of C++ online submissions for Search a 2D Matrix II.
Memory Usage: 13.1 MB, less than 6.67% of C++ online submissions for Search a 2D Matrix II.

三、优化措施

从左下角（右上角也行）出发，比较绝妙的解答：

class Solution{
	public:
		//从左下角（或者 右上角）出发，判断 
		bool searchMatrix(vector<vector<int> >& matrix,int target){
			if(matrix.size()<1){
				return false;
			}
			row = matrix.size();
			col = matrix[0].size();
			int curRow = row-1;
			int curCol = 0; 
			while(curRow<row && curCol<col){
				if(matrix[curRow][curCol] == target) return true;
				else if(matrix[curRow][curCol] < target){
					curCol ++;
				}else if(matrix[curRow][curCol] > target){
					curRow--;
				}
			}
			return false;
		} 
	private:
		int row,col;
};

性能如下：

Runtime: 72 ms, faster than 61.20% of C++ online submissions for Search a 2D Matrix II.
Memory Usage: 13 MB, less than 55.56% of C++ online submissions for Search a 2D Matrix II.

来源：https://www.cnblogs.com/siweihz/p/12292946.html