LeetCode 个人解答记录(1-5)

有待改进，本文章记录解题答案，以便日后查看以及改进

Two Sum

本题采用暴力求解法：

Runtime: 64 ms, faster than 40.98% of C++ online submissions for Two Sum.

vector<int> twoSum(vector<int>& nums, int target) {
    int n=nums.size();
    vector<int> result(2);
    for(int i=0;i<n;i++){
        int currentnum=nums[i];
        result[0]=i;
        for(int j=i+1;j<n;j++){
            if((currentnum+nums[j])==target){
                result[1]=j;
                return result;
            }
        }
    }
}

借助Map实现优化，map的KV为值以及值所对应的位置

Runtime: 8 ms, faster than 70.70% of C++ online submissions for Two Sum.

vector<int> twoSum(vector<int>& nums, int target) {
    int n=nums.size();
    map<int,int> numsMap;
    vector<int> result;
    for(int i=0;i<n;i++){
        numsMap[nums[i]]=i;
    }
    for(int i=0;i<n;i++){
        int res=target-nums[i];
        if(numsMap.find(res)!=numsMap.end()&&numsMap[res]!=i){
            result.push_back(i);
            result.push_back(numsMap[res]);
            return result;
        }
    }
}

Add Two Numbers

代码不够简洁，逻辑应该有地方可以合并,但是这样写速度很快

Runtime: 28 ms, faster than 98.46% of C++ online submissions for Add Two

ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
    ListNode* root =new ListNode(0); 
    ListNode* currentNode=root;
    int forward=0;
    while(l1!=NULL&&l2!=NULL){
        int current=l1->val+l2->val;
        currentNode->val=(forward+current)%10;
        forward=(forward+current-(forward+current)%10)/10;
        l1=l1->next;
        l2=l2->next;
        if(l1||l2){
            ListNode* nextNode=new ListNode(0);
            currentNode->next=nextNode;
            currentNode=nextNode;
        }    
    }
    
    while(l1!=NULL){
        int current=l1->val;
        currentNode->val=(forward+current)%10;
        forward=(forward+current-(forward+current)%10)/10;
        l1=l1->next;
        if(l1){
            ListNode* nextNode=new ListNode(0);
            currentNode->next=nextNode;
            currentNode=nextNode;
        }
    }
    
    while(l2!=NULL){
        int current=l2->val;
        currentNode->val=(forward+current)%10;
        forward=(forward+current-(forward+current)%10)/10;
        l2=l2->next;
        if(l2){
            ListNode* nextNode=new ListNode(0);
            currentNode->next=nextNode;
            currentNode=nextNode;
        }
    }
  
    if(forward!=0){
        ListNode* nextNode=new ListNode(forward);
        currentNode->next=nextNode;
        currentNode=nextNode;
    }
    
    return root;
}

Longest Substring Without Repeating Characters

采用Sliding window法，并且用map记录各个字母第一次出现所在位置来优化

Runtime: 24 ms, faster than 63.56% of C++ online submissions for Longest Substring Without Repeating Characters

int lengthOfLongestSubstring(string s) {
    map<char,int> mymap;
    int ans=0;
    int m=s.length();
    for(int i=0,j=0;j<m;j++){
        if(mymap.find(s[j])!= mymap.end()){
            i=max(mymap[s[j]],i);
        }
        ans=max(ans,j-i+1);
        mymap[s[j]]=j+1;
    }
    return ans;
}

Median of Two Sorted Arrays

找两个有序数组的中位数，时间复杂度 $O(log(m+n))$ ，之前腾讯面试问到，但是当时不会，这个解题思路还是很值得总结以及应用的。

思路：

在这里插入图片描述

在保证左右两堆数量是一样的情况下二分搜索切割A的方法，使得左边最大的小于右边最小的

Runtime: 52 ms, faster than 32.31% of C++ online submissions for Median of Two Sorted Arrays.

double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
    int m=nums1.size();
    int n=nums2.size();
    if(m>n){
        int tmp=m; m=n; n=tmp;
        vector<int> temp=nums2;
        nums2=nums1;
        nums1=temp;
    }
    int iMin=0;
    int iMax=m;
    int i,j=0;
    while(iMin<=iMax){
        i=(iMin+iMax)/2;
        j=(m+n+1)/2-i;
        if(i<iMax&&nums2[j-1]>nums1[i]){
            iMin=i+1;
        }else if(i>iMin&&nums1[i-1]>nums2[j]){
            iMax=i-1;
        }else{
            int maxLeft=0;
            if(i==0){
                maxLeft=nums2[j-1];
            }else if(j==0){
                maxLeft=nums1[i-1];
            }else{
                maxLeft=max(nums1[i-1],nums2[j-1]);
            }
            if((m+n)%2==1){
                return maxLeft;
            }
            int minRight=0;
            if(i==m){
                minRight=nums2[j];
            }else if(j==n){
                minRight=nums1[i];
            }else{
                minRight=min(nums1[i],nums2[j]);
            }
            cout<<maxLeft;
            cout<<minRight;
            double result=(maxLeft+minRight)/2.0;
            return result;
        }
    }
    return 0;
}

Longest Palindromic Substring

动态规划，区间DP

Runtime: 76 ms, faster than 38.20% of C++ online submissions for Longest Palindromic Substring.

string longestPalindrome(string s) {
    int n=s.length();
    
    int dp[1001][1001]={0};
    int result=0;
    int start=0;
    if(n>0){result=1;}
    for(int i=0;i<n;i++){
        dp[i][i]=1;
        if(s[i]==s[i+1]&&i+1<n){
            dp[i][i+1]=1;
            if(result<2){
                result=2;
                start=i;
            }
        }
        
    }

    for(int k=3;k<=n;k++){
        for(int i=0,j=i+k-1;j<n;i++,j++){
            if(dp[i+1][j-1]==1&&s[i]==s[j]){
                dp[i][j]=1;
                if(result<(j-i+1)){
                    result=j-i+1;
                    start=i;
                }
                
            }
        }
    }

    return s.substr(start,result);
}

LeetCode 个人解答记录(1-5)

LeetCode 个人解答记录(1-5)

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