You can not select more than 25 topics Topics must start with a letter or number, can include dashes ('-') and can be up to 35 characters long.

98 lines
2.0 KiB

This file contains ambiguous Unicode characters!

This file contains ambiguous Unicode characters that may be confused with others in your current locale. If your use case is intentional and legitimate, you can safely ignore this warning. Use the Escape button to highlight these characters.

package class14;
public class Code01_Parentheses {
public static boolean valid(String s) {
char[] str = s.toCharArray();
int count = 0;
for (int i = 0; i < str.length; i++) {
count += str[i] == '(' ? 1 : -1;
if (count < 0) {
return false;
}
}
return count == 0;
}
public static int needParentheses(String s) {
char[] str = s.toCharArray();
int count = 0;
int need = 0;
for (int i = 0; i < str.length; i++) {
if (str[i] == '(') {
count++;
} else { // 遇到的是')'
if (count == 0) {
need++;
} else {
count--;
}
}
}
return count + need;
}
public static boolean isValid(char[] str) {
if (str == null || str.length == 0) {
return false;
}
int status = 0;
for (int i = 0; i < str.length; i++) {
if (str[i] != ')' && str[i] != '(') {
return false;
}
if (str[i] == ')' && --status < 0) {
return false;
}
if (str[i] == '(') {
status++;
}
}
return status == 0;
}
public static int deep(String s) {
char[] str = s.toCharArray();
if (!isValid(str)) {
return 0;
}
int count = 0;
int max = 0;
for (int i = 0; i < str.length; i++) {
if (str[i] == '(') {
max = Math.max(max, ++count);
} else {
count--;
}
}
return max;
}
// s只由(和)组成
// 求最长有效括号子串长度
// 本题测试链接 : https://leetcode.com/problems/longest-valid-parentheses/
public static int longestValidParentheses(String s) {
if (s == null || s.length() < 2) {
return 0;
}
char[] str = s.toCharArray();
// dp[i] : 子串必须以i位置结尾的情况下往左最远能扩出多长的有效区域
int[] dp = new int[str.length];
// dp[0] = 0;
int pre = 0;
int ans = 0;
for (int i = 1; i < str.length; i++) {
if (str[i] == ')') {
// 当前谁和i位置的),去配!
pre = i - dp[i - 1] - 1; // 与str[i]配对的左括号的位置 pre
if (pre >= 0 && str[pre] == '(') {
dp[i] = dp[i - 1] + 2 + (pre > 0 ? dp[pre - 1] : 0);
}
}
ans = Math.max(ans, dp[i]);
}
return ans;
}
}