Given target =3, returntrue. class Solution { public: bool searchMatrix(vector<vector<int> > &matrix, int target) { if(matrix.size() == 0) return false; int m = matrix.size(); int n = matrix[0].size(); if(target < matrix[0][0] || target > matrix[m-1][n-1]) //目标...
Can you solve this real interview question? Search a 2D Matrix II - 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 in ascendin
class Solution: def searchMatrix(self, matrix: List[List[int]], target: int) -> bool: # 矩阵行数 m: int = len(matrix) # 矩阵列数 n: int = len(matrix[0]) # 二分区间左边界初始化为 0 l: int = 0 # 二分区间右边界初始化为 m * n - 1 r: int = m * n - 1 # 当二分...
方法一: 1classSolution {2public:3boolsearchMatrix(vector<vector<int> > &matrix,inttarget)4{5introw =matrix.size();6intcol = matrix[0].size();7intsubRow =0;8if(row ==0|| col ==0)returnfalse;910//寻找行11if(matrix[row -1][0] <= target)//最后一行,特殊处理12subRow = row -...
解法:虽然数组matrix中的每一行都是有序的,但是与Search a 2D matrix不同的是,不满足每行的首个元素比前一行的最后一个元素大;不过对于每一行还是有序的,所以,我们可以利用分治算法,对每一行利用二分查找; Java class Solution { public boolean searchMatrix(int[][] matrix, int target) { ...
1publicclassSolution {2privatebooleanIfFind =false;3privatevoidRightCornerRecur(int[][] matrix,inttarget,intm_start,intm_end,intn_start,intn_end){4if(m_start > m_end || n_start >n_end)5return;67if(m_start < 0 || n_start < 0 || m_end >= matrix.length || n_end >= matri...
Solution1 Code: classSolution1{publicbooleansearchMatrix(int[][]matrix,inttarget){if(matrix==null||matrix.length<1||matrix[0].length<1){returnfalse;}introw=0;intcol=matrix[0].length-1;while(col>=0&&row<=matrix.length-1){if(target==matrix[row][col]){returntrue;}elseif(target<matrix...
publicclassSolution{/* https://leetcode.com/discuss/48852/my-concise-o-m-n-java-solution */publicbooleansearchMatrix(int[][]matrix,inttarget){if(matrix==null||matrix.length<1||matrix[0].length<1){returnfalse;}introw=0,col=matrix[0].length-1;while(row<matrix.length&&col>=0){// star...
inthi=a.length()-1; while(lo<hi): intmi=lo+(hi-lo)>>1; if(a[mi]>target) lo=mi+1; else hi=mi; if(lo!=target) return-1; else returnlo; 3. 给定一个有序(非降序)数组A,可含有重复元素,求最大的i使得A[i]等于target,不存在则返回-1 ...
leetcode 240. Search a 2D Matrix II 思路:从左下或者右上开始进行搜索,因为比如从左下开始搜索,若目标数大于此时的数,接下来只能向右搜,若小于,接着只能向上搜。若是从左上开始进行搜索,若目标数大于此时的数,会有两个方向可以走。 ...240. Search a 2D Matrix II class Solution { public: bool ...