213 lines
6.3 KiB
Markdown
213 lines
6.3 KiB
Markdown
# 字符串循环移位包含
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[编程之美 3.1](#)
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```html
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s1 = AABCD, s2 = CDAA
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Return : true
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```
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给定两个字符串 s1 和 s2，要求判定 s2 是否能够被 s1 做循环移位得到的字符串包含。
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s1 进行循环移位的结果是 s1s1 的子字符串，因此只要判断 s2 是否是 s1s1 的子字符串即可。
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# 字符串循环移位
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[编程之美 2.17](#)
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```html
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s = "abcd123" k = 3
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Return "123abcd"
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```
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将字符串向右循环移动 k 位。
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将 abcd123 中的 abcd 和 123 单独翻转，得到 dcba321，然后对整个字符串进行翻转，得到 123abcd。
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# 字符串中单词的翻转
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[程序员代码面试指南](#)
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```html
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s = "I am a student"
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Return "student a am I"
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```
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将每个单词翻转，然后将整个字符串翻转。
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# 两个字符串包含的字符是否完全相同
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[242. Valid Anagram (Easy)](https://leetcode.com/problems/valid-anagram/description/)
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```html
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s = "anagram", t = "nagaram", return true.
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s = "rat", t = "car", return false.
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```
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可以用 HashMap 来映射字符与出现次数，然后比较两个字符串出现的字符数量是否相同。
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由于本题的字符串只包含 26 个小写字符，因此可以使用长度为 26 的整型数组对字符串出现的字符进行统计，不再使用 HashMap。
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```java
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public boolean isAnagram(String s, String t) {
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int[] cnts = new int[26];
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for (char c : s.toCharArray()) {
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cnts[c - 'a']++;
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}
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for (char c : t.toCharArray()) {
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cnts[c - 'a']--;
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}
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for (int cnt : cnts) {
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if (cnt != 0) {
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return false;
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}
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}
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return true;
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}
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```
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# 计算一组字符集合可以组成的回文字符串的最大长度
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[409. Longest Palindrome (Easy)](https://leetcode.com/problems/longest-palindrome/description/)
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```html
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Input : "abccccdd"
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Output : 7
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Explanation : One longest palindrome that can be built is "dccaccd", whose length is 7.
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```
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使用长度为 256 的整型数组来统计每个字符出现的个数，每个字符有偶数个可以用来构成回文字符串。
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因为回文字符串最中间的那个字符可以单独出现，所以如果有单独的字符就把它放到最中间。
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```java
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public int longestPalindrome(String s) {
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int[] cnts = new int[256];
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for (char c : s.toCharArray()) {
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cnts[c]++;
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}
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int palindrome = 0;
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for (int cnt : cnts) {
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palindrome += (cnt / 2) * 2;
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}
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if (palindrome < s.length()) {
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palindrome++; // 这个条件下 s 中一定有单个未使用的字符存在，可以把这个字符放到回文的最中间
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}
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return palindrome;
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}
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```
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# 字符串同构
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[205. Isomorphic Strings (Easy)](https://leetcode.com/problems/isomorphic-strings/description/)
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```html
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Given "egg", "add", return true.
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Given "foo", "bar", return false.
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Given "paper", "title", return true.
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```
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记录一个字符上次出现的位置，如果两个字符串中的字符上次出现的位置一样，那么就属于同构。
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```java
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public boolean isIsomorphic(String s, String t) {
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int[] preIndexOfS = new int[256];
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int[] preIndexOfT = new int[256];
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for (int i = 0; i < s.length(); i++) {
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char sc = s.charAt(i), tc = t.charAt(i);
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if (preIndexOfS[sc] != preIndexOfT[tc]) {
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return false;
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}
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preIndexOfS[sc] = i + 1;
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preIndexOfT[tc] = i + 1;
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}
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return true;
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}
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```
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# 回文子字符串个数
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[647. Palindromic Substrings (Medium)](https://leetcode.com/problems/palindromic-substrings/description/)
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```html
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Input: "aaa"
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Output: 6
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Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa".
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```
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从字符串的某一位开始，尝试着去扩展子字符串。
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```java
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private int cnt = 0;
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public int countSubstrings(String s) {
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for (int i = 0; i < s.length(); i++) {
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extendSubstrings(s, i, i); // 奇数长度
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extendSubstrings(s, i, i + 1); // 偶数长度
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}
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return cnt;
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}
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private void extendSubstrings(String s, int start, int end) {
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while (start >= 0 && end < s.length() && s.charAt(start) == s.charAt(end)) {
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start--;
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end++;
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cnt++;
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}
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}
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```
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# 判断一个整数是否是回文数
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[9. Palindrome Number (Easy)](https://leetcode.com/problems/palindrome-number/description/)
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要求不能使用额外空间，也就不能将整数转换为字符串进行判断。
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将整数分成左右两部分，右边那部分需要转置，然后判断这两部分是否相等。
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```java
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public boolean isPalindrome(int x) {
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if (x == 0) {
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return true;
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}
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if (x < 0 || x % 10 == 0) {
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return false;
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}
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int right = 0;
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while (x > right) {
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right = right * 10 + x % 10;
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x /= 10;
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}
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return x == right || x == right / 10;
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}
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```
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# 统计二进制字符串中连续 1 和连续 0 数量相同的子字符串个数
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[696. Count Binary Substrings (Easy)](https://leetcode.com/problems/count-binary-substrings/description/)
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```html
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Input: "00110011"
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Output: 6
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Explanation: There are 6 substrings that have equal number of consecutive 1's and 0's: "0011", "01", "1100", "10", "0011", and "01".
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```
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```java
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public int countBinarySubstrings(String s) {
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int preLen = 0, curLen = 1, count = 0;
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for (int i = 1; i < s.length(); i++) {
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if (s.charAt(i) == s.charAt(i - 1)) {
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curLen++;
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} else {
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preLen = curLen;
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curLen = 1;
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}
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if (preLen >= curLen) {
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count++;
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}
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}
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return count;
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}
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```
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