• 经典算法(四) 数组相关 & 螺旋矩阵 & 数字大小写转换 & 字符串相关


    一、求所有子数组的和的最大值

    public static void main(String[] args) {
            int[] a = { 1, -2, 3, 10, -4, 7, 2, -5 };
            FindMaxSubAry(a);
        }
    
        public static void FindMaxSubAry(int[] arr) {
            int maxStartIndex = 0;// 最大数组开始坐标
            int maxEndIndex = 0; // 最大数组结束坐标
            int max = 0; // 最大数组的和
            int tempSum = 0;
            for (int i = 0; i < arr.length; i++) {
                tempSum += arr[i];
                if (tempSum <= 0) {
                    tempSum = 0;
                    max = 0;
                    maxStartIndex = i + 1;
                }
                if (tempSum > max) {
                    max = tempSum;
                    maxEndIndex = i;
                }
            }
            System.out.println(" sum:" + max);
            print(arr, maxStartIndex, maxEndIndex);
        }
    
        public static void print(int[] arr, int startIndex, int endIndex) {
            for (int i = startIndex; i <= endIndex; i++) {
                System.out.print(arr[i] + " ");
            }
            System.out.println();
        }
    View Code

    二、顺时针打印矩阵

       分析:构造二维矩阵

       1     2    3     4    
       5     6    7     8    
       9    10   11   12   
      13  14   15   16  
       按照从外到里以顺时针的顺序依次打印
       打印:1 2 3 4 8 12 16 15 14 13 9 5 6 7 11 10

    public static void print(int[][] a,int start,int end){
            if(start>=end||end<0){
                return;
            }
            for(int i=start;i<=end;i++){
                System.out.print(a[start][i]+" ");
            }
    
            for(int i=start+1;i<=end;i++){
                System.out.print(a[i][end]+" ");
            }
            for(int i=end-1;i>=start;i--){
                System.out.print(a[end][i]+" ");
            }
    
            for(int i=end-1;i>start;i--){
                System.out.print(a[i][start]+" ");
            }
            print(a, start+1, end-1);
        }
    View Code

     三、2110980789转化为“二十一亿一千零九十八万零七百八十九”

    分析:可以分4位为一个量级。其中0789 量级单位 为空,1098量级单位为 万,21量级单位为亿。这四位字符串在以个十百千的形式计算。

    package com.amei.java.algorithm;
    
    import java.util.HashMap;
    import java.util.Map;
    import java.util.Stack;
    /**
     * 将阿拉伯数字转换为大写数字
     * 123010980789转化为“二十一亿一千零九十八万零七百八十九”
     * @param num
     */
    public class Common {
        static Map<Integer, String> map = new HashMap<Integer, String>();
        static {
            map.put(1, "");
            map.put(2, "十");
            map.put(3, "百");
            map.put(4, "千");
        }
    
        public static void main(String[] args) {
            NumberChange(2110980789);
        }
    
        static String change(int k, char ch) {
            String number = numTobigString(ch);
            String unit = map.get(k);
            if (ch == '0') {
                return number;
            } else {
                return number + unit;
            }
        }
    
        /**
         * 123010980789转化为“二十一亿一千零九十八万零七百八十九”
         * 
         * @param num
         */
        static void NumberChange(int num) {
            String s = num + "";
    
            char[] ch = s.toCharArray();
            int flag = 0;//控制量级单位
            int k = 1;
            Stack<String> st = new Stack<String>();
             for (int i = ch.length - 1; i >= 0; i--) {
                if (flag == 1 && k == 1) {
                    st.add("万");
                }
                if (flag == 2 && k == 1) {
                    st.add("亿");
                }
                st.add(change(k, ch[i]));
                
                if (k % 4 == 0) {
                    flag++;
                    k = 1;
                } else {
                    k++;
                }
            }
            int size = st.size();
            for (int i = 0; i < size; i++) {
                System.out.print(st.pop());
            }
        }
    
        static String numTobigString(char c) {
            if (c == '1') {
                return "一";
            } else if (c == '2') {
                return "二";
            } else if (c == '3') {
                return "三";
            } else if (c == '4') {
                return "四";
            } else if (c == '5') {
                return "五";
            } else if (c == '6') {
                return "六";
            } else if (c == '7') {
                return "七";
            } else if (c == '8') {
                return "八";
            } else if (c == '9') {
                return "九";
            } else if (c == '0') {
                return "零";
            }
            return null;
        }
    }
    View Code

     四、判断一个ip是否属于某个ip段

    分析:将ip转换为数字比较大小,可以看成ip是逢256就进位的数字

    public static boolean ipExistsInRange(String ip, String ipSection) {
            ipSection = ipSection.trim();
            ip = ip.trim();
            int idx = ipSection.indexOf('-');
            String beginIP = ipSection.substring(0, idx);
            String endIP = ipSection.substring(idx + 1);
            return getIp2long2(beginIP) <= getIp2long2(ip) && getIp2long2(ip) <= getIp2long2(endIP);
        }
    
        public static long getIp2long(String ip) {
            ip = ip.trim();
            String[] ips = ip.split("\.");
            long ip2long = 0L;
            for (int i = 0; i < 4; ++i) {
                //左移8位,相当于乘以256
                ip2long = ip2long << 8 | Integer.parseInt(ips[i]);
            }
            return ip2long;
        }
    
        public static long getIp2long2(String ip) {
            ip = ip.trim();
            String[] ips = ip.split("\.");
            long ip1 = Integer.parseInt(ips[0]);
            long ip2 = Integer.parseInt(ips[1]);
            long ip3 = Integer.parseInt(ips[2]);
            long ip4 = Integer.parseInt(ips[3]);
            long ip2long = 1L * ip1 * 256 * 256 * 256 + ip2 * 256 * 256 + ip3 * 256 + ip4;
            return ip2long;
        }
    
        public static void main(String[] args) {
            String ip = "10.10.10.116";
            String ipSection = "10.10.1.00-10.10.255.255";
            boolean exists = ipExistsInRange(ip, ipSection);
            System.out.println(exists);
        }
    View Code

     五、将字符串 abcd;3456789; 反转为 dcba;9876543;

    public class StrReverse {
        public static void main(String[] args) {
            String str = "abcd;3456789";
            System.out.println(sectionReverse(str));
        }
    
        static String sectionReverse(String str) {
            char[] c = str.toCharArray();
            int start = 0;
            int end = 0;
            for (int i = 0; i < c.length; i++) {
                if (c[i] == ';') {
                    end = i - 1;
                    reverse(c, start, end);
                    start = i + 1;
                }else if(i==c.length-1){
                    end=i;
                    reverse(c, start, end);
                }
            }
            return String.valueOf(c);
        }
    
        static void reverse(char[] c, int start, int end) {
            char temp;
            for (int j = 0; j < (end - start) / 2 + 1; j++) {
                temp = c[end - j];
                c[end - j] = c[j + start];
                c[j + start] = temp;
            }
        }
    }
    View Code

     六、统计字符串“abaabbccdfeffsfsgggaabgee”中 出现且仅出现n次的第一个字母

    public class StrRepeat {
    
        public static void main(String[] args) {
            String str = "abaabbccdfeffsfsgggaabgee;";
            System.out.println(repeatCharFisrt(str, 4));
        }
    
        static char repeatCharFisrt(String str, int repeatNum) {
            Map<Character, Integer> map = new LinkedHashMap<Character, Integer>();
            char[] c = str.toCharArray();
            for (int i = 0; i < c.length; i++) {
                if (map.containsKey(c[i])) {
                    map.put(c[i], map.get(c[i]) + 1);
                } else {
                    map.put(c[i], 1);
                }
            }
            for (java.util.Map.Entry<Character, Integer> entry : map.entrySet()) {
                if (entry.getValue() == repeatNum) {
                    return entry.getKey();
                }
            }
            return '0';
        }
    }
    View Code


     

  • 相关阅读:
    记一次主从同步出现错误
    Mycat的学习
    MHA高可用集群
    MySQL进行 行累计
    设定从某个时间执行脚本,直到现在
    跟踪数据
    爬虫笔记
    爬虫练习
    css3-文本阴影
    vue生命周期
  • 原文地址:https://www.cnblogs.com/amei0/p/8336804.html
Copyright © 2020-2023  润新知