LeetCode 301. Remove Invalid Parentheses（删除无效的括号）

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原题网址：https://leetcode.com/problems/remove-invalid-parentheses/

Remove the minimum number of invalid parentheses in order to make the input string valid. Return all possible results.

Note: The input string may contain letters other than the parentheses ( and ).

Examples:

"()())()" -> ["()()()", "(())()"]"(a)())()" -> ["(a)()()", "(a())()"]")(" -> [""]

方法一：广度优先搜索。

public class Solution {    public List<String> removeInvalidParentheses(String s) {        Set<String> visit = new HashSet<>();        List<String> results = new ArrayList<>();        LinkedList<String> queue = new LinkedList<>();        queue.add(s);        boolean succ = false;        while (!queue.isEmpty()) {            String q = queue.remove();            if (succ) match(q, results); else succ = match(q, visit, queue, results);        }        return results;    }        private void match(String s, List<String> results) {        // System.out.printf("Try matching %s\n", s);        int match = 0;        for(int i=0; i<s.length(); i++) {            if (s.charAt(i) == '(') match ++;            else if (s.charAt(i) == ')') match --;            if (match < 0) return;        }        if (match == 0) results.add(s);        // System.out.printf("Try matching %s successfully\n", s);    }        private boolean match(String s, Set<String> visit, LinkedList<String> queue, List<String> results) {        // System.out.printf("Try matching %s\n", s);        int match = 0;        for(int i = 0; i < s.length(); i ++) {            if (s.charAt(i) == '(') match ++;            else if (s.charAt(i) == ')') match --;            if (match < 0) {                for(int j=0; j<=i; j++) {                    if (s.charAt(j) != ')') continue;                    String r = s.substring(0, j) + s.substring(j+1);                    if (visit.contains(r)) continue;                    visit.add(r);                    queue.add(r);                }                // System.out.printf("Try matching %s fail, queue=%s\n", s, queue);                return false;            }        }        if (match == 0) {            results.add(s);            // System.out.printf("Try matching %s successfully\n", s);            return true;        }        match = 0;        for(int i = s.length()-1; i >= 0; i --) {            if (s.charAt(i) == ')') match ++;            else if (s.charAt(i) == '(') match --;            if (match < 0) {                for(int j=s.length()-1; j>=i; j--) {                    if (s.charAt(j) != '(') continue;                    String r = s.substring(0, j) + s.substring(j+1);                    if (visit.contains(r)) continue;                    visit.add(r);                    queue.add(r);                }                // System.out.printf("Try matching %s fail, queue=%s\n", s, queue);                return false;            }        }        // System.out.printf("Try matching %s fail, queue=%s\n", s, queue);        return false;    }}

方法二：深度优先搜索。

public class Solution {    private Set<String> visit = new HashSet<>();    private List<String> results = new ArrayList<>();        private void removeLeft(String s) {        int match = 0;        for(int i=s.length()-1; i>=0; i--) {            if (s.charAt(i) == ')') match ++;            else if (s.charAt(i) == '(') match --;            if (match < 0) {                for(int j=s.length()-1; j>=i; j--) {                    if (s.charAt(j) != '(') continue;                    String r = s.substring(0, j) + s.substring(j+1);                    if (visit.contains(r)) continue;                    visit.add(r);                    removeLeft(r);                }                return;            }        }        results.add(s);    }    private void removeRight(String s) {        int match = 0;        for(int i=0; i<s.length(); i++) {            if (s.charAt(i) == '(') match ++;            else if (s.charAt(i) == ')') match --;            if (match < 0) {                for(int j=0; j<=i; j++) {                    if (s.charAt(j) != ')') continue;                    String r = s.substring(0, j) + s.substring(j+1);                    if (visit.contains(r)) continue;                    visit.add(r);                    removeRight(r);                }                return;            }        }        if (match == 0) {            results.add(s);            return;        }        removeLeft(s);    }    public List<String> removeInvalidParentheses(String s) {        removeRight(s);        return results;    }}

方法三：先删除无效的右括号，再删除无效的左括号，深度优先搜索。

这道题我没有做出好的解法，参考：http://algobox.org/remove-invalid-parentheses/

精妙之处在于：使用lastj保存上一次发现不匹配的位置和上一次删除括号的位置。

但如何证明唯一性？这个我还没有搞懂。

public class Solution {    private void removeLeft(String s, int lastI, int lastJ, List<String> rs) {        int match = 0;        for(int i=lastI; i<s.length() && i>=0; i--) {            if (s.charAt(i) == ')') match ++;            else if (s.charAt(i) == '(') match --;            if (match >= 0) continue;            for(int j=lastJ; j<s.length() && j>=i; j--) {                if (s.charAt(j) != '(') continue;                if (j==lastJ || s.charAt(j+1) != '(') removeLeft(s.substring(0, j) + s.substring(j+1, s.length()), i-1, j-1, rs);            }            return;        }        rs.add(s);    }    private void removeRight(String s, int lastI, int lastJ, List<String> rs) {        int match = 0;        for(int i=lastI; i<s.length(); i++) {            if (s.charAt(i) == '(') match ++;            else if (s.charAt(i) == ')') match --;            if (match >= 0) continue;            for(int j=lastJ; j<=i; j++) {                if (s.charAt(j) != ')') continue;                if (j==lastJ || s.charAt(j-1) != ')') removeRight(s.substring(0, j) + s.substring(j+1, s.length()), i, j, rs);            }            return;        }        removeLeft(s, s.length()-1, s.length()-1, rs);    }    public List<String> removeInvalidParentheses(String s) {        List<String> results = new ArrayList<>();        removeRight(s, 0, 0, results);        return results;    }}

带注释：

public class Solution {    private void removeLeft(String s, int matchTo, int removeTo, List<String> results) {        int matched = 0;        for(int m=matchTo; m>=0; m--) {            if (s.charAt(m) == ')') matched ++;            else if (s.charAt(m) == '(') matched --;            if (matched >= 0) continue;            for(int r=removeTo; r>=m; r--) {                //检查是左括号才能删除                if (s.charAt(r) != '(') continue;                if (r==removeTo || s.charAt(r+1) != '(') removeLeft(s.substring(0, r)+s.substring(r+1), m-1, r-1, results);            }            //如果本次有删除，则留待后面加入到results            return;        }        results.add(s);    }    private void removeRight(String s, int matchTo, int removeTo, List<String> results) {        int matched = 0;        for(int m=matchTo; m<s.length(); m++) {            if (s.charAt(m) == '(') matched ++;            else if (s.charAt(m) == ')') matched --;            if (matched >= 0) continue;            for(int r=removeTo; r<=m; r++) {                //检查是右括号才能删除                if (s.charAt(r) != ')') continue;                if (r==removeTo || s.charAt(r-1) != ')') removeRight(s.substring(0, r)+s.substring(r+1), m, r, results);            }            //如果本次有删除，则留待后面加入到results            return;        }        removeLeft(s, s.length()-1, s.length()-1, results);    }    public List<String> removeInvalidParentheses(String s) {        List<String> results = new ArrayList<>();        removeRight(s, 0, 0, results);        return results;    }}

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