而为了能够方便在given 2D matrix找到需要比对的值,我们还是需要确定行数和列数,通过上表可以看出,行数是position/columns,而列数是position%columns, 这样一来,就能很容易的在原矩阵中定位到所需要的值。剩下其他的解题思路,就与二分查找法一模一样了。 时间复杂度O(log(rows*columns)) 代码如下: 1publicboolea...
publicbooleansearchMatrix(int[][] matrix,inttarget) { if(matrix.length==0){ returnfalse; } intcol=matrix[0].length;//数组列数 introw=matrix.length;//数组行数 int[] newArray=newint[row*col];//将二维数组放到一个一维素中,newArray[i]对应二维数组中的matrix[i/n][i%n] returnbinarySearch...
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]) //目标元素不在矩阵中 return false; int low = 0, high = m ...
所以根据matrix[i][j]和target的大小关系, 要么往左走, 要么往下走 */ class Solution { public boolean searchMatrix(int[][] matrix, int target) { if(matrix==null || matrix.length==0 || matrix[0].length==0) return false; boolean flag = false; int rows = matrix.length, cols = matrix...
来自专栏 · LeetCode 每日一题 题意 给定一个 m * n 的整数矩阵 matrix ,每一行从左到右升序,每一行的第一个数字比上一行的最后一个数字大,判断 target 是否在这个矩阵中? 数据限制 m == matrix.length n == matrix[i].length 1 <= m, n <= 100 -(10 ^ 4) <= matrix[i][j], target <...
reference:https://discuss.leetcode.com/topic/33240/java-an-easy-to-understand-divide-and-conquer-method 思路: zone1zone2***|***|***|***|***---***|***|***|***|***zone3zone4 First, we divide the matrix into four quarters as shown below: We then compare ...
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...
Can you solve this real interview question? Search a 2D Matrix - You are given an m x n integer matrix matrix with the following two properties: * Each row is sorted in non-decreasing order. * The first integer of each row is greater than the last int
2019-11-24 12:03 −原题链接在这里:https://leetcode.com/problems/search-in-a-sorted-array-of-unknown-size/ 题目: Given an integer array sorted in ascending order, write a fun... Dylan_Java_NYC 0 1310 [LC] 81. Search in Rotated Sorted Array II ...
# 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, #...