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[0];
        for(int i=1;i<strs.size();i++)
        {
            comm = "";
            for(int j=0;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 (2,-1) ;
        if(n<=0) return pos;
        //left bound
        int start=0,stop = n-1,mid;
        while(start<=stop)
        {
            mid = (start+stop)/2;
            if(A[mid]==target && (mid==0 || A[mid-1]<target))
            {
                pos[0] = mid;
                break;
            }
            else if(A[mid]<target)
                 {
                     start = mid+1;
                 }
                 else{
                     stop = mid-1;
                 }
        }
        if(pos[0]==-1) return pos;
        //right bound
        start=0,stop = n-1;
        while(start<=stop)
        {
            mid = (start+stop)/2;   
            if(A[mid]==target && (mid==n-1 || A[mid+1]>target) )
            {
                pos[1] = mid;
                break;
            }
            else if(A[mid]>target)
                 {
                     stop = mid-1;
                 }
                 else{
                     start = mid+1;
                 }
        }
        return pos;
    }
};