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(); }
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); }
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; } }
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); }
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; } } }
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'; } }
