标签：des c style class blog code
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,布布扣,bubuko.com
Leetcode::Longest Common Prefix && Search for a Range
标签：des c style class blog code
原文地址：http://www.cnblogs.com/soyscut/p/3776985.html
