a 字符串被包含在 b 字符串中
示例 1：
输入：s = "abc", t = "ahbgdc"
输出：true
示例 2：
输入：s = "axc", t = "ahbgdc"
输出：false
@Test
public void test() {
String s = "abc", t = "ahbgdc";
int sPoint = 0,tPoint = 0;
while (sPoint < s.length() && tPoint<t.length()) {
if (s.charAt(sPoint) == t.charAt(tPoint)){
sPoint++;
}
tPoint++;
}
System.out.println(sPoint== s.length());
}
判断s 是t 的连续子串
@Test
public void test() {
String s = "abc", t = "ahabcgdc";
int sPoint = 0, tPoint = 0;
while (tPoint < t.length()) {
if (s.charAt(sPoint) == t.charAt(tPoint)) {
sPoint++;
if (sPoint == s.length()) {
break;
}
} else {
sPoint = 0;
}
tPoint++;
}
System.out.println(sPoint == s.length());
}
字符串左移右移
@Test
public void moveLeft(){
String xx = "abcd123";
int k = 3;
char[] cs = new char[256];
int index = 0;
for (int i = 0; i < xx.length(); i++) {
cs[index++] = xx.charAt((i+xx.length()+k)%xx.length());
}
for (int i = 0; i < index; i++) {
System.out.println(cs[i]);
}
}
@Test
public void moveRight(){
String xx = "abcd123";
int k = 3;
char[] cs = new char[256];
int index = 0;
for (int i = 0; i < xx.length(); i++) {
cs[index++] = xx.charAt((i+xx.length()-k)%xx.length());
}
for (int i = 0; i < index; i++) {
System.out.println(cs[i]);
}
}
判断两次字符串的字符完全相同
@Test
public void test(){
String s = "anagram", t = "nagaram";
String s1 = "rat", t1 = "car";
System.out.println(isSame(s,t));
System.out.println(isSame(s1,t1));
}
public boolean isSame(String s,String t) {
if (s.length() != t.length()) {
return false;
}
int[] check = new int[26];
for (int i = 0; i < t.length(); i++) {
check[s.charAt(i)-'a'] ++;
check[t.charAt(i)-'a'] --;
}
for (int i : check) {
if (i != 0) {
return false;
}
}
return true;
}
计算一组字符集合可以组成的回文字符串的最大长度
@Test
public void test(){
String xx = "abccaa";
System.out.println(LongestPalindrome(xx));
}
public int LongestPalindrome(String x) {
char[] charArray = x.toCharArray();
int[] xc = new int[26];
for (char c : charArray) {
xc[c-'a'] ++;
}
int res = 0;
for (int i = 0; i < xc.length; i++) {
res += (xc[i] /2) *2;
}
return res == charArray.length?res:res+1;
}
第一个不重复的字符
@Test
public void call() {
String xx = "google";
System.out.println(test(xx));
}
public int test(String xx){
int[] xs = new int[26];
for (int i = 0; i < xx.length(); i++) {
xs[xx.charAt(i)-'a'] ++;
}
for (int i = 0; i < xx.length(); i++) {
if (xs[xx.charAt(i)-'a'] == 1) {
return i;
}
}
return -1;
}