一、Easy 696 Count Binary Substrings
Input: "00110011"
Output: 6
Explanation: There are 6 substrings that have equal number of consecutive 1's and 0's: "0011", "01", "1100", "10", "0011", and "01".
Input: "10101"
Output: 4
Explanation: There are 4 substrings: "10", "01", "10", "01" that have equal number of consecutive 1's and 0's.
解题思路：由于字符串是0和1相互交叉的，因此：
1.将s = "110001111000000"分成四个子字符串：group = {11 000 1111 000000
2.如果group[i] = 2, group[i+1]=3,那么两个组合起来，一定有11000或者是00111
3.类似于000111的情况，那么就有‘0’*3+‘1’*3，‘0’*2+‘1’*2，‘0’*1+‘1’*1，一共三种情况，因此如果group[i]和group[j]中的最小长度就是可以构成的个数
解题代码：
（1）(O(N) + O(1)) 注意groups[++t] = 1;和groups[t]++;的表示技巧。
public int countBinarySubstrings(String s) {
int[] groups = new int[s.length()];
int t = 0;
groups[0] = 1;
for (int i = 1; i < s.length(); i++) {
if (s.charAt(i-1) != s.charAt(i)) {
groups[++t] = 1;
} else {
groups[t]++;
}
}
int ans = 0;
for (int i = 1; i <= t; i++) ans += Math.min(groups[i-1], groups[i])；
return ans;
}
（2）不用group数组而是直接使用两个int类型的数prev和curr来存储每个部分的长度
class Solution {
public int countBinarySubstrings(String s) {
int ans = 0, prev = 0, cur = 1;
for (int i = 1; i < s.length(); i++) {
if (s.charAt(i-1) != s.charAt(i)) {
ans += Math.min(prev, cur);
prev = cur;
cur = 1;
} else {
cur++;
}
}
return ans + Math.min(prev, cur);
}
}
二、Easy 686. Repeated String Match
For example, with A = "abcd" and B = "cdabcdab".
Return 3, because by repeating A three times (“abcdabcdabcd”),
B is a substring of it; and B is not a substring of A repeated two times ("abcdabcd").
1.解法思路
求出满足A*k（k=1,2,3...n）的子字符串是B的最小的k值。
2.解法代码
时间复杂度O(N∗(N+M))
空间复杂度O(M+N)
需要学习的地方：①使用StringBuilder可以一次性append一个字符串。
②for ( ; sb.length() < B.length(); number++) sb.append(A);注意for循环的这种使用技巧与number++的位置。
③使用sb.indexOf(B) >= 0和sb.append(A).indexOf(B) >= 0结合sb.length() < B.length()条件很方便地判断出结果来。
public int repeatedStringMatch(String A, String B) {
int number = 1;
StringBuilder sb = new StringBuilder(A);
for ( ; sb.length() < B.length(); number++) sb.append(A);
if (sb.indexOf(B) >= 0) return number;
if (sb.append(A).indexOf(B) >= 0) return number+1;
return -1;
}
三、680. Valid Palindrome II
Given a non-empty string s, you may delete at most one character. Judge whether you can make it a palindrome.
1.正常算法Brute Force [Time Limit Exceeded]
public boolean validPalindrome(String s) {
for (int i = 0; i < s.length(); i++) {
StringBuilder sb = new StringBuilder(s.substring(0, i).concat(s.substring(i+1)));
String result = sb.toString();
if (result.equals(sb.reverse().toString())) return true;
}
return false;
}
或者
public boolean isPalindrome(CharSequence s) {
for (int i = 0; i < s.length() / 2; i++) {
if (s.charAt(i) != s.charAt(s.length() - 1 - i)) {
return false;
}
}
return true;
}
public boolean validPalindrome(String s) {
StringBuilder sb = new StringBuilder(s);
for (int i = 0; i < s.length(); i++) {
char c = sb.charAt(i);
sb.deleteCharAt(i);
if (isPalindrome(sb)) return true;
sb.insert(i, c);
}
return isPalindrome(s);
}
2.贪心算法Greedy[Accepted]（TODO）
四、67. Add Binary
Input: a = "1010", b = "1011"
Output: "10101"
1.传统方法
public String addBinary(String a, String b) {
String longStr = (a.length() > b.length()) ? a : b;
String shortStr = (a.length() > b.length()) ? b : a;
StringBuilder sb = new StringBuilder();
boolean isCarry = false;
for (int i = shortStr.length()-1; i >= 0; i--) {
if (isCarry) {
if ((longStr.charAt(longStr.length()-shortStr.length()+i) == '0' && shortStr.charAt(i) == '0')) {
sb.append('1');
isCarry = false;
} else if (longStr.charAt(longStr.length()-shortStr.length()+i) == '0' && shortStr.charAt(i) == '1') {
sb.append('0');
isCarry = true;
} else if (longStr.charAt(longStr.length()-shortStr.length()+i) == '1' && shortStr.charAt(i) == '0') {
sb.append('0');
isCarry = true;
} else {
sb.append('1');
isCarry = true;
}
} else {
if ((longStr.charAt(longStr.length()-shortStr.length()+i) == '0' && shortStr.charAt(i) == '0')) {
sb.append('0');
} else if (longStr.charAt(longStr.length()-shortStr.length()+i) == '0' && shortStr.charAt(i) == '1') {
sb.append('1');
} else if (longStr.charAt(longStr.length()-shortStr.length()+i) == '1' && shortStr.charAt(i) == '0') {
sb.append('1');
} else {
sb.append('0');
isCarry = true;
}
}
}
for (int i = longStr.length() - shortStr.length() - 1; i >= 0; i--) {
if (isCarry) {
if (longStr.charAt(i) == '1') {
sb.append('0');
isCarry = true;
} else {
sb.append('1');
isCarry = false;
}
} else {
sb.append(longStr.charAt(i));
}
}
if (isCarry) sb.append('1');
return sb.reverse().toString();
}
传统方法
2.非传统数学方法（TODO）（或者是使用堆栈）
public String addBinary(String a, String b) {
StringBuilder sb = new StringBuilder();
int i = a.length() - 1, j = b.length() -1, carry = 0;
while (i >= 0 || j >= 0) {
int sum = carry;
if (j >= 0) sum += b.charAt(j--) - '0';
if (i >= 0) sum += a.charAt(i--) - '0';
sb.append(sum % 2);
carry = sum / 2;
}
if (carry != 0) sb.append(carry);
return sb.reverse().toString();
}
五、