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字符串算法

字符串算法

作者: 万福来 | 来源:发表于2020-06-02 19:19 被阅读0次

    最长公共前缀

    class Solution {
        public String longestCommonPrefix(String[] strs) {
            if(strs == null || strs.length == 0)
                return "";
            Arrays.sort(strs);
            char [] first = strs[0].toCharArray();
            char [] last = strs[strs.length - 1].toCharArray();
            StringBuffer res = new StringBuffer();
            int len = first.length < last.length ? first.length : last.length;
            int i = 0;
            while(i < len){
                if(first[i] == last[i]){
                    res.append(first[i]);
                    i++;
                }
                else
                    break;
            }
            return res.toString();
        }
    }
    

    最长回文串

    class Solution {
        //最长回文串
        public int longestPalindrome(String s) {
            HashSet<Character> hs = new HashSet<>();
            int len = s.length();
            int count = 0;
            if(len == 0)
                return 0;
            for(int i = 0; i<len; i++){
                if(hs.contains(s.charAt(i))){
                    hs.remove(s.charAt(i));
                    count++;
                }else{
                    hs.add(s.charAt(i));
                }
            }
            return hs.isEmpty() ? count * 2 : count * 2 + 1;
        }
    }
    

    最长回文子序列

    class Solution {
        //最长回文子序列
        public int longestPalindromeSubseq(String s) {
            int len = s.length();
            int [][] dp = new int[len][len];
            for(int i = len - 1; i>=0; i--){
                dp[i][i] = 1;
                for(int j = i+1; j < len; j++){
                    if(s.charAt(i) == s.charAt(j))
                        dp[i][j] = dp[i+1][j-1] + 2;
                    else
                        dp[i][j] = Math.max(dp[i+1][j], dp[i][j-1]);
                }
            }
            return dp[0][len-1];
        }
    }
    

    最长公共子串

    func SerachMaxCommString(arr1 ,arr2 []int) []int {
        len1 := len(arr1)
        len2 := len(arr2)
        //利用二维数组记录子问题的LCS长度
        opt := make([][]int,len1,len1)
        for i := 0; i < len1; i++ {
            opt[i] = make([]int, len2)
        }
        for i:=0;i<len1;i++{
            for j:=0;j<len2;j++{
                if arr1[i] == arr2[j] {
                    if i == 0 || j == 0 {
                        opt[i][j] = 1
                    }else {
                        opt[i][j] = opt[i-1][j-1] + 1
                    }
                }else{
                    opt[i][j] = 0
                }
            }
        }
        //找出最大公共长度和坐标
        max := opt[0][0]
        indexi := 0
        indexj := 0
        for i:=0;i<len1;i++{
            for j:=0;j<len2;j++{
                if opt[i][j] > max {
                    max = opt[i][j]
                    indexi = i
                    indexj = j
                }
            }
        }
        fmt.Printf("max lcs = %v \n",max)
        fmt.Printf("max lcs indexi= %v \n",indexi)
        fmt.Printf("max lcs indexj = %v \n",indexj)
        //取短的串返回
        narr := make([]int,max)
        index := 0
        var shortArr []int
        if len1 < len2 {
            index = indexi
            shortArr = arr1
        }else {
            index = indexj
            shortArr = arr2
        }
        fmt.Println(index)
        fmt.Println(shortArr)
        for i:=0;i<max;i++{
            fmt.Println(shortArr[index-max+1+i])
            narr[i] = shortArr[index-max+1+i]
        }
        return narr
    }
    

    反转单词顺序列

    public String reverseWords(String s) {
            if(s.trim().length() == 0)
                return s.trim();
            String [] temp = s.trim().split(" +");
            String res = "";
            for(int i = temp.length - 1; i > 0; i--){
                res += temp[i] + " ";
            }
            return res + temp[0];
        }
    

    反转字符串

    public String reverseString(String s) {
            if(s.length() < 2)
                return s;
            int l = 0, r = s.length() - 1;
            char [] strs = s.toCharArray(); 
            while(l < r){
                char temp = strs[l];
                strs[l] = strs[r];
                strs[r] = temp;
                l++;
                r--;
            }
            return new String(strs);
        }
    

    字符串转数字

    public class Solution {
        public int StrToInt(String str) {
            if(str.length() == 0)
                return 0;
            int flag = 0;
            if(str.charAt(0) == '+')
                flag = 1;
            else if(str.charAt(0) == '-')
                flag = 2;
            int start = flag > 0 ? 1 : 0;
            long res = 0;
            while(start < str.length()){
                if(str.charAt(start) > '9' || str.charAt(start) < '0')
                    return 0;
                res = res * 10 + (str.charAt(start) - '0');
                start ++;
            }
            return flag == 2 ? -(int)res : (int)res;
        }
    }
    
    

    IP-int互转

    func IpToInt(ip string) (int32, error) {
        ipArr := strings.Split(ip, ".")
        if len(ipArr) != 4 {
            return 0, errors.New("IP is wrong")
        }
        ip1, err1 := strconv.Atoi(ipArr[0])
        ip2, err2 := strconv.Atoi(ipArr[1])
        ip3, err3 := strconv.Atoi(ipArr[2])
        ip4, err4 := strconv.Atoi(ipArr[3])
        if err1 != nil || err2 != nil || err3 != nil || err4 != nil {
            log.Println(err1, err2, err3, err4)
            return 0, errors.New("toint error")
        }
        ipInt := ip1<<24 + ip2<<16 + ip3<<8 + ip4
        return int32(ipInt), nil
    }
    func IntToIp(ip int32) string {
        ip1 := ip >> 24 & 0xFF
        ip2 := ip >> 16 & 0xFF
        ip3 := ip >> 8 & 0xFF
        ip4 := ip & 0xFF
        ips := fmt.Sprintf("%d.%d.%d.%d", ip1, ip2, ip3, ip4)
        return ips
    }
    

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