携程笔试87%+100%+0%，顺便求第3道答案
1. 链表排序(87%)：不知道哪里没考虑到，没ac
static ListNode partition(ListNode head,int m) {
if(head == null) {
return null;
}
boolean first = true;
ListNode insertNode = head, cur = head, prev = null;
while(cur != null) {
if(cur.val <= m && cur == head) { // 链表头即满足(不需要移动)
first = false;
prev = cur;
cur = cur.next;
}else if(cur.val <= m && first){ // 非链表头，第一次满足条件需要移动
first = false;
prev.next = cur.next;
cur.next = head;
head = cur;
insertNode = cur;
cur = prev.next;
}else if(cur.val <= m) { // 非链表头，非首次满足条件需要移动
prev.next = cur.next;
cur.next = insertNode.next;
insertNode.next =cur;
insertNode = cur;
cur = prev.next;
}else { // 不满足条件，下一个
prev = cur;
cur = cur.next;
}
}
return head;
} 2. 表达式解析(100%)：比较明显的栈使用场景 static String resolve(String expr) {
if(expr == null || expr == "") {
return "";
}
Stack<Character> stack = new Stack<>();
char[] c = expr.toCharArray();
int len = c.length;
if(len == 1) {
if(c[0] == '(' || c[0] == ')') {
return "";
}else {
return expr;
}
}
String res = "";
for(int i=0;i<len;i++) {
if(c[i] == ')') { // 需要弹栈
String s = "";
while(stack.peek() != '(') {
s += stack.pop();
}
if(stack.peek() != '(') {
return "";
}else {
stack.pop();
}
if(stack.isEmpty()) {
res += s;
}else {
for(int j=0;j<s.length();j++) {
stack.push(s.charAt(j));
}
}
}else if(c[i] == '(' || !stack.isEmpty()){
stack.push(c[i]);
}else{
res += c[i];
}
}
return stack.isEmpty() ? res : "";
} 3.国际惯例第3道时间不够，没想出来也，大佬们有写出来的说说吧😅
