131. Palindrome Partitioning

刷
July-14-2019

没秒了，不应该啊。。想复杂了，没必要带着当前string进入下一层，直接加到TEMPLIST里
abc =>
dfs([a], bc) -> dfs([ab], c) -> dfs([abc], c)
dfs([a,b], c) -> dfs([a, bc], "")

Time: O(2^n)

    // T(n)   = T(n-1) + T(n-2) + T(n-3) ... + T(1)
    // T(n-1) =          T(n-2) + T(n-3) ... + T(1)
    // T(n) = 2T(n-1)

class Solution {
    public List<List<String>> partition(String s) {
        List<List<String>> res = new ArrayList<>();
        if (s == null || s.length() == 0) return res;
        
        dfs(res, s, new ArrayList<>());
        
        return res;
    }
    
    // T(n)   = T(n-1) + T(n-2) + T(n-3) ... + T(1)
    // T(n-1) =          T(n-2) + T(n-3) ... + T(1)
    // T(n) = 2T(n-1)
    public void dfs(List<List<String>> res, String s, List<String> tempList) {
        if (s.length() == 0) {
            res.add(new ArrayList<>(tempList));
        } else {
            for (int i = 1; i <= s.length(); i ++) {
                String leftStr = s.substring(0, i);
                if (!isPalindrome(leftStr)) continue;
                tempList.add(leftStr);
                dfs(res, s.substring(i), tempList);
                tempList.remove(tempList.size() - 1);
            }

        }
    }
    
    public boolean isPalindrome(String s) {
        int l = 0, r = s.length() - 1;
        while (l < r) {
            if (s.charAt(l ++) != s.charAt(r --)) {
                return false;
            }
        }   
        return true;
    }
}

---

三刷。

DFS + BACKTRACK

T(n) = T(n - 1) + T(n-2) + ..+ T(n-1)
T(n) = 2T(n-1) = 4T(n-2) = 2^n T(1)
isValid: O(n)
Time: O(n*2^n)
space: O(2^n) 跟一个STRING能有多少种分法一样。。

public class Solution {
    public List<List<String>> partition(String s) {
        List<List<String>> res = new ArrayList<List<String>>();
        if (s.length() == 0) return res;
        dfs(res, s, new ArrayList<>());
        return res;
    }
    
    public void dfs(List<List<String>> res, String s, List<String> tempList) {
        if (s.length() == 0) {
            res.add(tempList);
            return;
        } else {
            for (int i = 1; i <= s.length(); i++) {
                String str1 = new String(s.substring(0,i));
                if (!isValid(str1)) continue;
                String str2 = new String(s.substring(i));
                tempList.add(str1);
                dfs(res, str2, new ArrayList<>(tempList));
                tempList.remove(tempList.size() - 1);
            }
            return;
        }
    }
    
    public boolean isValid(String s) {
        if (s.length() <= 1) return true;
        int l = 0;
        int r = s.length() - 1;
        while (l < r) {
            if (s.charAt(l++) != s.charAt(r--)) return false;
        }
        return true;
    }
}

以前做出来自鸣得意，现在回头看，难点从来都不是题本身，而是时间和空间的复杂度…