一次总结两道题，两道题目都比较基础
Description:Write a function to find the longest common prefix string
amongst an array of strings.
分析： 这道题目最重要的知道什么叫prefix前缀， 否则一不小心就做成了最长子序列。前缀就是两个序列的前多少位
都要是一样的，不能跳着对齐，这样就比较简单了，也可以求很多个序列的公共前缀。
1 class Solution {
2 public:
3 string longestCommonPrefix(vector<string> &strs) {
4 string result,comm;
5 if(strs.empty()) return result;
6 result= strs[0];
7 for(int i=1;i<strs.size();i++)
8 {
9 comm = "";
10 for(int j=0;j<result.size();j++)
11 {
12 if(j>=strs[i].size()) break;
13 if(result[j]==strs[i][j]) comm.insert(comm.end(),result[j]);
14 else break;
15 }
16 result = comm;
17 }
18
19 return result;
20 }
21 };
题目二：Search for a range
Description: Given a sorted array of integers, find the starting and ending
position of a given target value.
Your algorithm‘s runtime complexity must be in the order
of O(log n).
If the target is not found in the array, return [-1,.
-1]
For example,
Given [5, 7, 7, 8, 8, 10] and target
value 8,
return [3, 4].
分析： 要找一个排序数组的某个元素的范围，要求复杂度是O(log n),则肯定是基于二分查找。因为有重复元素，则将二分查找
的边界条件变一下就可以了。
1 class Solution {
2 public:
3 vector<int> searchRange(int A[], int n, int target) {
4 vector<int> pos (2,-1) ;
5 if(n<=0) return pos;
6 //left bound
7 int start=0,stop = n-1,mid;
8 while(start<=stop)
9 {
10 mid = (start+stop)/2;
11 if(A[mid]==target && (mid==0 || A[mid-1]<target))
12 {
13 pos[0] = mid;
14 break;
15 }
16 else if(A[mid]<target)
17 {
18 start = mid+1;
19 }
20 else{
21 stop = mid-1;
22 }
23 }
24 if(pos[0]==-1) return pos;
25 //right bound
26 start=0,stop = n-1;
27 while(start<=stop)
28 {
29 mid = (start+stop)/2;
30 if(A[mid]==target && (mid==n-1 || A[mid+1]>target) )
31 {
32 pos[1] = mid;
33 break;
34 }
35 else if(A[mid]>target)
36 {
37 stop = mid-1;
38 }
39 else{
40 start = mid+1;
41 }
42 }
43 return pos;
44 }
45 };
Leetcode::Longest Common Prefix && Search for a
Range
时间： 2024-10-10 06:00:47