Contents of this section
Warm Reminder/Opening Introduction/Speaking Before Brushing Up
Starting today, I started to write questions, I believe this is a long process!
This column is the classic topic of "Jianzhi offer",
And record your understanding and learning process in this blog !
Topic introduction (LINK)
As for the topic, I won’t introduce too much, it’s not very meaningful to copy it, so I’ll just post the link here!
LeetCode link: click here!
NowCoder link: click here!
own thoughts/thoughts
1. Initial Thoughts / Final Thoughts
In fact, my idea is to search while traversing the two-dimensional array , which is relatively simple and simple!
However, in fact, what the interviewer wants to see is not this ,
Because anyone can think of such a simple logic!
The feature of the topic is not even used: sorting in increasing order!
A good idea is this:
- The essence of the search process is the process of exclusion. It is best if you can exclude a batch!
Look at the picture below:
Finally, there is a critical question to consider : when will it end?
In the end, if it is not found, it means that the condition of a row or column out of bounds leads to dissatisfaction!
2. Notes
If you see the word "whether or not " in the title , you can judge that the return value is a Boolean type!
Code for my first attempt
I didn't write it out at the beginning, so I got confused!
Then I wrote it after a simple understanding (this is the so-called violent law, which is meaningless):
public class Solution {
public boolean Find(int target, int [][] array) {
//数组为空或者长度为0直接返回false
if (array == null || array.length == 0) {
return false;
}
for (int i = 0; i < array.length; i++) {
for (int j = 0; j < array[0].length; j++) {
if (array[i][j] == target) {
return true;
}
}
}
return false;
}
}
code after research
class Solution {
public boolean findNumberIn2DArray(int[][] matrix, int target) {
//如果为空或者长度为0，直接返回false
if (matrix == null || matrix.length == 0) {
return false;
}
int i = 0;//列下标
int j = matrix[0].length - 1;//行下标，第一行的长度-1
while (i < matrix.length && j >= 0) {
//这个就是整个查找的范围
//范围就是：
if (target < matrix[i][j]) {
j--;//排除列
}else if (target > matrix[i][j]) {
i++;//排除行
} else {
return true;
}
}
return false;
}
}
作者：cbiltps
链接：https://leetcode-cn.com/problems/er-wei-shu-zu-zhong-de-cha-zhao-lcof/solution/yong-you-shang-jiao-de-zhi-bi-jiao-ji-ke-gz3e/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
Thank you
When the first question comes to the question, I just pull the switch, I didn’t make it, and I just read the answer...
a bit lost...