class Solution {
public String decodeString(String s) {
StringBuilder res = new StringBuilder();
int multi = 0;
LinkedList<Integer> stack_multi = new LinkedList<>();
LinkedList<String> stack_res = new LinkedList<>();
for(Character c : s.toCharArray()) {
if(c == '[') {
stack_multi.addLast(multi);
stack_res.addLast(res.toString());
multi = 0;
res = new StringBuilder();
} else if(c == ']') {
StringBuilder tmp = new StringBuilder();
int cur_multi = stack_multi.removeLast();
for(int i = 0; i < cur_multi; i++) tmp.append(res);
res = new StringBuilder(stack_res.removeLast() + tmp);
} else if(c >= '0' && c <= '9')
multi = multi * 10 + Integer.parseInt(c + "");
else res.append(c);
}
return res.toString();
}
}
class Solution {
public String decodeString(String s) {
return dfs(s, 0)[0];
}
private String[] dfs(String s, int i) {
StringBuilder res = new StringBuilder();
int multi = 0;
while(i < s.length()) {
if(s.charAt(i) >= '0' && s.charAt(i) <= '9')
multi = multi * 10 + Integer.parseInt(String.valueOf(s.charAt(i)));
else if(s.charAt(i) == '[') {
String[] tmp = dfs(s, i + 1);
i = Integer.parseInt(tmp[0]);
while(multi > 0) {
res.append(tmp[1]);
multi--;
}
}
else if(s.charAt(i) == ']')
return new String[] { String.valueOf(i), res.toString() };
else
res.append(String.valueOf(s.charAt(i)));
i++;
}
return new String[] { res.toString() };
}
}
class Solution {
public int convertInteger(int A, int B) {
int c = A^B;
int count = 0;
while(c!=0){
if((c&1)==1){
count++;
}
c >>>= 1;
}
return count;
}
}
// n&(n - 1)可以去掉一個數的二進制表示的最右邊的1
class Solution {
public int convertInteger(int A, int B) {
int num = A^B;
int count = 0;
while (num != 0) {
num &= num - 1;
count++;
}
return count;
}
}
class Solution {
public int reverseBits(int num) {
int maxLen = 0, preLen = 0, curLen = 0, bits = 32;
while (bits-- > 0) {
if ((num & 1) == 0) {
curLen -= preLen;
preLen = curLen + 1;
}
curLen++;
maxLen = Math.max(maxLen, curLen);
num >>= 1;
}
return maxLen;
}
}
你知道的越多,你不知道的越多。