leetcode

Posted 2021-01-31 也很简单

leetcode

• schedule: 3 be done ,3 better than 99%

  1.sum

Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

---

my code

思路：先对所给序列进行排序，然后依次对小于target/2的数a，在有序序列中用二分查找找(target-a)

• 复杂度
  排序调用标准库的sort，O(nlogn)
  二分查找O(n)次*单次O(logn)=O(nlogn)
  共O(nlong)+O(nlogn)=O(nlogn)

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> ivec=nums;
        sort(ivec.begin(),ivec.end());
        int h=target/2;//half
        int n1,n2,i;
        for(i=0;i<nums.size();i++)
        {
            if(ivec[i]<h)
            {
                if(binary_search(ivec.begin(),ivec.end(),target-ivec[i]))
                    break;
            }              
            else if(ivec[i]==h)
            {
                if(target%2==0){//偶数
                    auto r=equal_range(ivec.begin(),ivec.end(),h);
                    if(r.second-r.first>1)
                        break;
                }
                else//奇数
                {
                    break;
                }
            }
        }
        n1=ivec[i];
        n2=target-n1;
        vector<int>::iterator it1,it2;
        it1=find(nums.begin(),nums.end(),n1);
        if(n1==n2)
            it2=find(it1+1,nums.end(),n2);
        else
            it2=find(nums.begin(),nums.end(),n2);
        int p1=it1-nums.begin(),p2=it2-nums.begin();
        vector<int> result={p1,p2};
        return result;
    }
};

---

someone else‘s better

• 思路：同样，先排序。但排序后处理更为巧妙。时间复杂度仍为O(nlogn)

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> nums1(nums);
        sort(nums1.begin(),nums1.end());
        vector<int> re;
        int L = 0, R = nums.size()-1;
        while(L<=R){
            if(nums1[L]+nums1[R]<target)
                L++;
            else if(nums1[L]+nums1[R]>target)
                R--;
            else
                break;
        }
        int len = nums.size();
        for(int i = 0;i<len;i++){
            if(nums[i]==nums1[L]||nums[i]==nums1[R])
                re.push_back(i);
        }
        return re;
    }
};

---

官方解法：hash_map，时间复杂度O(n)

2.Add Two Numbers

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

---

my code (almost same as reference version)

• 思路：实际上和我们做加法手算一样，从低位开始一次像高位计算即可。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */

static int x=[](){//IO加速
    std::ios::sync_with_stdio(false);
    cin.tie(NULL);
    return 0;
}();

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* ls=l1,*lsp;//sum,sum‘s parent 
        int sum,c(0),t1,t2;
        while(l2!=NULL || ls!=NULL || c!=0)
        {
            t1=(ls==NULL)?0:ls->val;
            t2=(l2==NULL)?0:l2->val;
            sum=t1+t2+c;
            if(ls==NULL) 
            {
                ListNode* s=new ListNode(sum%10);
                ls=s;
                lsp->next=s;
            }
            else
                ls->val=sum%10;
            c=sum/10;
            lsp=ls;
            if(ls!=NULL) ls=ls->next; 
            if(l2!=NULL) l2=l2->next; 
        }    
        return l1;
    }
};

3.Longest Substring Without Repeating Characters

• 思路：用数组l[128]存储各个字符上一次出现的位置，maxLoad存储当前已出现的无重复字符子串的最大长度，i为当前考虑的子串的起点，j为当前考虑的子串的重点。
  迭代过程：不断前移j，并视情况移动i。
  参考答案称之为 Sliding Window ，行吧。
• 时间复杂度，O(n)

static int x=[](){//IO加速
    std::ios::sync_with_stdio(false);
    cin.tie(NULL);
    return 0;
}();


class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        int i,j,maxLong;//i:当前子串起点，j:当前子串终点
        int l[128];
        for(i=0;i<128;l[i++]=-1);
        for(i=0,j=0,maxLong=0;j<s.size();j++)
        {
            if(l[s[j]]!=-1 && l[s[j]]+1>i) i=l[s[j]]+1;
            l[s[j]]=j;
            if(j-i+1>maxLong) maxLong=j-i+1;
        }
        return maxLong;
    }
};