Leetcode::Longest Common Prefix && Search for a Range

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。

 class Solution {
 public:
     string longestCommonPrefix(vector<string> &strs) {
         string result,comm;
         if(strs.empty()) return result;
         result= strs[];
         for(int i=;i<strs.size();i++)
         {
             comm = "";
             for(int j=;j<result.size();j++)
             {
                 if(j>=strs[i].size()) break;
                 if(result[j]==strs[i][j]) comm.insert(comm.end(),result[j]);
                 else break;
             }
             result = comm;
         }

         return result;
     }
 };

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。

 class Solution {
 public:
     vector<int> searchRange(int A[], int n, int target) {
         vector<int> pos (,-) ;
         if(n<=) return pos;
         //left bound
         int start=,stop = n-,mid;
         while(start<=stop)
         {
             mid = (start+stop)/;
             if(A[mid]==target && (mid== || A[mid-]<target))
             {
                 pos[] = mid;
                 break;
             }
             else if(A[mid]<target)
                  {
                      start = mid+;
                  }
                  else{
                      stop = mid-;
                  }
         }
         if(pos[]==-) return pos;
         //right bound
         start=,stop = n-;
         while(start<=stop)
         {
             mid = (start+stop)/;
             if(A[mid]==target && (mid==n- || A[mid+]>target) )
             {
                 pos[] = mid;
                 break;
             }
             else if(A[mid]>target)
                  {
                      stop = mid-;
                  }
                  else{
                      start = mid+;
                  }
         }
         return pos;
     }
 };