一. 删除字符串中所有相邻重复项
class Solution {
public String removeDuplicates(String s) {
StringBuffer ret = new StringBuffer();
char[] ss = s.toCharArray();
for (int i = 0; i < ss.length; i++) {
int len = ret.length();
if (len > 0 && ss[i] == ret.charAt(len - 1)) {
ret.deleteCharAt(len - 1);
} else {
ret.append(ss[i]);
}
}
return ret.toString();
}
}
二. 比较含退格的字符串
class Solution {
public boolean backspaceCompare(String s, String t) {
return changeStr(s).equals(changeStr(t));
}
public String changeStr(String s){
StringBuffer ret = new StringBuffer();
char[] ss = s.toCharArray();
for(char x : ss){
int len = ret.length();
if(x == '#'){
if(len > 0){
ret.deleteCharAt(len - 1);
}
}else{
ret.append(x);
}
}
return ret.toString();
}
}
三. 基本计算器II
属于表达式求值的问题, 解法都是用栈模拟计算过程
class Solution {
public int calculate(String s) {
Deque<Integer> ret = new ArrayDeque<>();
char[] ss = s.toCharArray();
int n = ss.length;
int i = 0;
char op = '+';
while(i < n){
//判断空格
if(ss[i] == ' '){
i++;
}else if(ss[i] >= '0' && ss[i] <= '9'){
//如果是数字
int tmp = 0;
while(i < n && ss[i] >= '0' && ss[i] <= '9'){
//把数字提出来
tmp = tmp * 10 + (ss[i]-'0');
i++;
}
//如果+-就入栈, 如果*/ 就先与栈顶元素计算后入栈
if(op == '+') ret.push(tmp);
if(op == '-') ret.push(-tmp);
if(op == '*') ret.push(ret.poll() * tmp);
if(op == '/') ret.push(ret.poll() / tmp);
}else{
//如果是符号
op = ss[i];
i++;
}
}
//将栈中的元素全部相加
int sum = 0;
while(!ret.isEmpty()){
sum += ret.poll();
}
return sum;
}
}
四. 字符串解码
class Solution {
public String decodeString(String s) {
//一个存放字符串, 一个存放数字
Stack<StringBuffer> st = new Stack<>();
Stack<Integer> nums = new Stack<>();
//细节: 先放入一个空串
st.push(new StringBuffer());
int i = 0;
char[] ss = s.toCharArray();
int n = ss.length;
while(i < n){
//如果是数字, 就把这个数提出来, 放到数字栈
if(ss[i] >= '0' && ss[i] <= '9'){
int tmp = 0;
while(i < n && ss[i] >= '0' && ss[i] <= '9'){
tmp = tmp * 10 + (ss[i] - '0');
i++;
}
nums.push(tmp);
}else if(ss[i] == '['){
//如果是[, 就把后面的字符串提出来, 放到字符串栈
i++;
StringBuffer tmp = new StringBuffer();
while(i < n && ss[i] >= 'a' && ss[i] <= 'z'){
tmp.append(ss[i]);
i++;
}
st.push(tmp);
}else if(ss[i] ==']'){
//如果是], 就进行解析, 取两个栈的栈顶元素, 添加到字符串栈栈顶的后面
StringBuffer str = st.pop();
int num = nums.pop();
while(num-- != 0){
st.peek().append(str);
}
i++;
}else{
//如果是不在[]中的字符串, 就直接添加到字符串栈栈顶的后面
StringBuffer tmp = new StringBuffer();
while(i < n && ss[i] >= 'a' && ss[i] <= 'z'){
tmp.append(ss[i]);
i++;
}
st.peek().append(tmp);
}
}
return st.pop().toString();
}
}
五. 验证栈序列
class Solution {
public boolean validateStackSequences(int[] pushed, int[] popped) {
Stack<Integer> st = new Stack<>();
int j = 0;
for(int i = 0; i < pushed.length; i++){
st.push(pushed[i]);
while(!st.isEmpty() && popped[j] == st.peek()){
st.pop();
j++;
}
}
if(st.isEmpty()){
return true;
}
return false;
}
}