leetcode.水题*3

Posted 2020-09-08 此剑之势愈斩愈烈

Longest Substring Without Repeating Characters   

找无相同字符的最长子串，o(n)暴力判断即可

class Solution {
public:
	int a[300];
    int lengthOfLongestSubstring(string s) {
        int len=s.length();
        if(len==0)  return 0;
        memset(a,0,sizeof(a));
        int ans=1;
        int sum=1;
        a[s[0]]=1;
        for(int i=1;i<len;i++)
    	{
    		int shit=s[i];
    		if(sum+1>i-a[s[i]]+1)
    			sum=i-a[s[i]]+1;
    		else
    			sum++;
    	//	cout<<sum<<" "<<i-a[s[i]]+1<<endl;
    		if(sum>ans)	ans=sum;
    		a[shit]=i+1;
		}
		return ans;
    }
};

16. 3Sum Closest

对于给定整数，在给定数组中寻找三个数的和，使得这个和最接近这个整数

显然二分即可

class Solution {
public:
	int mx= 0x7fffffff;
	int bs(vector<int>& a,int left,int right,int tar)
	{
		int x=left,y=right;
		while(left<=right)
		{
			int mid=(left+right)>>1;
			if(a[mid]>tar)
				right=mid-1;
			else
				left= mid+1;
		}
		//cout<<left-1;
		int shit,fuck;
		shit=fuck=mx;
		if(left-1>=x)
		shit=tar-a[left-1];
		if(left<=y)
		fuck=a[left]-tar;
		if(shit<fuck)
			return left-1;
		else
			return left;
	}
	int fb(int x)
	{
	    if(x>0)
	        return x;
	    return -x;
	}
    int threeSumClosest(vector<int>& nums, int target) {
        sort(nums.begin(),nums.end());
        int len=nums.size();
        int ans=mx;
        int x=0;
        for(int i=0;i<len;i++)
        	x+=nums[i];
        for(int i=0;i<len;i++)
        {
        	int sum=nums[i];
        	for(int j=i+1;j<len-1;j++)
        	{
        		sum+=nums[j];
        		int shit=bs(nums,j+1,len-1,target-sum);
        		shit=nums[shit];
        	//	cout<<ans<<endl;
        		if(ans>fb(target-shit-sum))
        		{
        			ans=fb(target-shit-sum);
        			x=shit+sum;
        		}
        		sum-=nums[j];
			}
		}
		return x;
    }
};

 14. Longest Common Prefix

找一系列字符串的公共子串，直接暴力

 

class Solution {
public:
    string longestCommonPrefix(vector<string>& strs) {
        int len=strs.size();
        if(len==0)
        {
            string ans="";
            return ans;
        }
        string ans=strs[0];
        int j;
        for(j=0;j<ans.length();j++)
        {   
            int i;
            for(i=1;i<len;i++)
                if(ans[j]!=strs[i][j])
                    break;
            if(i<len)
            {
                break;
            }
        }
        ans=ans.substr(0,j);
        return ans;
    }
};