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Copy path1249.minimum-remove-to-make-valid-parentheses.java
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50 lines (44 loc) · 1.51 KB
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/*
* @lc app=leetcode id=1249 lang=java
*
* [1249] Minimum Remove to Make Valid Parentheses
*/
// @lc code=start
class Solution {
public String minRemoveToMakeValid(String s) {
if (s == null || s.length() == 0) {
return s;
}
StringBuilder sb = new StringBuilder();
int leftParCnt = 0;
for (char currChar : s.toCharArray()) {
// 如果是左括号,左括号记数加一
if (currChar == '(') {
leftParCnt++;
}
else if (currChar == ')') {
// 如果当前是右括号但是没有左括号去抵消,则需要忽略这个右括号
if (leftParCnt == 0) {
continue;
}
// 否则左括号记数减一,代表合并了
leftParCnt--;
}
// 把当前的char放进sb里
sb.append(currChar);
}
// 上面循环的结果,一定是左括号不少于右括号
StringBuilder res = new StringBuilder();
for (int i = sb.length() - 1; i >= 0; i--) {
char currChar = sb.charAt(i);
// 如果当前的左括号是多余的没法合并的,因为可以合并的都在上面那个循环里合并掉了
// 则跳过当前的左括号
if (currChar == '(' && leftParCnt-- > 0) {
continue;
}
res.append(currChar);
}
return res.reverse().toString();
}
}
// @lc code=end