Hard模式题目

先过一下Hard模式的题目吧。

	#	Title	Editorial	Acceptance	Difficulty	Frequency
。	65	Valid Number		12.6%	Hard
。	126	Word Ladder II		13.6%	Hard
。	149	Max Points on a Line		15.6%	Hard
。	146	LRU Cache		16.0%	Hard
。	68	Text Justification		18.1%	Hard
。	460	LFU Cache		19.0%	Hard
。	44	Wildcard Matching		19.0%	Hard
。	308	Range Sum Query 2D - Mutable		19.8%	Hard
。	4	Median of Two Sorted Arrays		20.8%	Hard
。	420	Strong Password Checker		21.0%	Hard
。	273	Integer to English Words		21.1%	Hard
。	30	Substring with Concatenation of All Words		21.7%	Hard
。	440	K-th Smallest in Lexicographical Order		22.1%	Hard
。	140	Word Break II		22.2%	Hard
。	212	Word Search II		22.2%	Hard
。	269	Alien Dictionary		22.3%	Hard
。	174	Dungeon Game		22.9%	Hard
。	214	Shortest Palindrome		23.0%	Hard
	446	Arithmetic Slices II - Subsequence		23.0%	Hard
	32	Longest Valid Parentheses		23.1%	Hard
	295	Find Median from Data Stream		23.3%	Hard
	132	Palindrome Partitioning II		23.4%	Hard
	10	Regular Expression Matching		23.6%	Hard
	76	Minimum Window Substring		23.8%	Hard
	188	Best Time to Buy and Sell Stock IV		23.8%	Hard
	321	Create Maximum Number		23.9%	Hard
	135	Candy		23.9%	Hard
	335	Self Crossing		23.9%	Hard
	97	Interleaving String		23.9%	Hard
	391	Perfect Rectangle		24.2%	Hard
	158	Read N Characters Given Read4 II - Call multiple times		24.4%	Hard
	466	Count The Repetitions		24.6%	Hard
	336	Palindrome Pairs		24.7%	Hard
	41	First Missing Positive		24.9%	Hard
	124	Binary Tree Maximum Path Sum		25.0%	Hard
	224	Basic Calculator		25.5%	Hard
	218	The Skyline Problem		25.5%	Hard
	84	Largest Rectangle in Histogram		25.6%	Hard
	23	Merge k Sorted Lists		25.9%	Hard
	45	Jump Game II		26.0%	Hard
	85	Maximal Rectangle		26.1%	Hard
	57	Insert Interval		26.3%	Hard
	138	Copy List with Random Pointer		26.6%	Hard
	233	Number of Digit One		27.3%	Hard
	381	Insert Delete GetRandom O(1) - Duplicates allowed		28.0%	Hard
	37	Sudoku Solver		28.1%	Hard
	432	All O`one Data Structure		28.2%	Hard
	87	Scramble String		28.3%	Hard
	123	Best Time to Buy and Sell Stock III		28.3%	Hard
	56	Merge Intervals		28.4%	Hard
	282	Expression Add Operators		28.5%	Hard
	316	Remove Duplicate Letters		28.6%	Hard
	164	Maximum Gap		28.6%	Hard
	99	Recover Binary Search Tree		28.7%	Hard
	327	Count of Range Sum		28.9%	Hard
	51	N-Queens		29.0%	Hard
	25	Reverse Nodes in k-Group		29.7%	Hard
	472	Concatenated Words		30.1%	Hard
	465	Optimal Account Balancing		30.1%	Hard
	248	Strobogrammatic Number III		30.5%	Hard
	72	Edit Distance		30.6%	Hard
	115	Distinct Subsequences		30.6%	Hard
	403	Frog Jump		30.9%	Hard
	411	Minimum Unique Word Abbreviation		31.1%	Hard
	239	Sliding Window Maximum		31.4%	Hard
	330	Patching Array		31.5%	Hard
	297	Serialize and Deserialize Binary Tree		31.6%	Hard
	354	Russian Doll Envelopes		31.8%	Hard
	358	Rearrange String k Distance Apart		31.8%	Hard
	33	Search in Rotated Sorted Array		31.9%	Hard
	363	Max Sum of Rectangle No Larger Than K		32.1%	Hard
	410	Split Array Largest Sum		32.2%	Hard
	480	Sliding Window Median		33.1%	Hard
	317	Shortest Distance from All Buildings		33.3%	Hard
	117	Populating Next Right Pointers in Each Node II		33.5%	Hard
	315	Count of Smaller Numbers After Self		33.5%	Hard
	301	Remove Invalid Parentheses		34.5%	Hard
	42	Trapping Rain Water		35.3%	Hard
	128	Longest Consecutive Sequence		35.3%	Hard
	329	Longest Increasing Path in a Matrix		35.4%	Hard
	407	Trapping Rain Water II		35.6%	Hard
	154	Find Minimum in Rotated Sorted Array II		36.2%	Hard
	265	Paint House II		37.1%	Hard
	272	Closest Binary Search Tree Value II		37.6%	Hard
	291	Word Pattern II		37.7%	Hard
	305	Number of Islands II		38.1%	Hard
	380	Insert Delete GetRandom O(1)		38.4%	Hard
	145	Binary Tree Postorder Traversal		38.4%	Hard
	340	Longest Substring with At Most K Distinct Characters		38.6%	Hard
	352	Data Stream as Disjoint Intervals		38.9%	Hard
	159	Longest Substring with At Most Two Distinct Characters		39.7%	Hard
	312	Burst Balloons		41.6%	Hard
	287	Find the Duplicate Number		41.8%	Hard
	425	Word Squares		41.9%	Hard
	52	N-Queens II		42.8%	Hard
	302	Smallest Rectangle Enclosing Black Pixels		44.0%	Hard
	471	Encode String with Shortest Length		44.2%	Hard
	296	Best Meeting Point		50.4%	Hard

65

Valid Number

12.6%

Hard

好像很繁琐。

126

Word Ladder II

13.6%

Hard

方法是很好的。不是采用查找备选集的方式，而是采用逐个字符替换(可以加Hash来查找)，这样复杂度从很高，降低到O(26*n), n是字典长度。

149

Max Points on a Line

15.6%

Hard

解法很好。通过角度来判断是否一条直线。我看了解法之后，觉得可以再优化一下，比如记下后续点已经处理过的角度。

146

LRU Cache

16.0%

Hard

我觉得用一个Hash, 加一个双向链表比较好。Hash里面记录链表实际节点，然后每次访问，这个节点移到双向链表的头部，成为新的head。每次满了删除，就删除尾部的。

里面还用了stl的list实现里面的一种很快的方式（当然了，自己来移动节点，也行）：

// 貌似用这个splice比较快
dl.splice(dl.begin(), dl, mitr->second);

68

Text Justification

18.1%

Hard

这道题目主要的难点就是gap的处理，尤其是gap不一致的情况。用的是下面的方法：

if (i < extra) {
line.push_back(' ');
}

也就是说，对于前面"extra"个数的间隔，都加上一个。

460

LFU Cache

19.0%

Hard

很好，非常好！用了一个抽象数据结构Node，来记录同一访问频次层级的key，然后用了LinkedHashSet来记录这些key，进一步保证了先来后到（先达到的，会先删除）。好好领悟，很好！

还用了两个HashMap，分别记录真实的结果，以及一个key对应的Node（如上所述，这个Node里面包含了很多key，用LinkedHashSet来组织的）

44

Wildcard Matching

19.0%

Hard

下面这个真是写的非常非常好：

C++版本的：

bool isMatch(const char *s, const char *p) {
        const char* star=NULL;
        const char* ss=s;
        while (*s){
            //advancing both pointers when (both characters match) or ('?' found in pattern)
            //note that *p will not advance beyond its length
            if ((*p=='?')||(*p==*s)){s++;p++;continue;} 

            // * found in pattern, track index of *, only advancing pattern pointer
            if (*p=='*'){star=p++; ss=s;continue;} 

            //current characters didn't match, last pattern pointer was *, current pattern pointer is not *
            //only advancing pattern pointer
            if (star){ p = star+; s=++ss;continue;} 

           //current pattern pointer is not star, last patter pointer was not *
           //characters do not match
            return false;
        }

       //check for remaining characters in pattern
        while (*p=='*'){p++;}

        return !*p;
    }

Java版本的：

boolean comparison(String str, String pattern) {
        int s = 0, p = 0, match = 0, starIdx = -1;
        while (s < str.length()){
            // advancing both pointers
            if (p < pattern.length()  && (pattern.charAt(p) == '?' || str.charAt(s) == pattern.charAt(p))){
                s++;
                p++;
            }
            // * found, only advancing pattern pointer
            else if (p < pattern.length() && pattern.charAt(p) == '*'){
                starIdx = p;
                match = s;
                p++;
            }
           // last pattern pointer was *, advancing string pointer
            else if (starIdx != -1){
                p = starIdx + 1;
                match++;
                s = match;
            }
           //current pattern pointer is not star, last patter pointer was not *
          //characters do not match
            else return false;
        }

        //check for remaining characters in pattern
        while (p < pattern.length() && pattern.charAt(p) == '*')
            p++;

        return p == pattern.length();
}

4

Median of Two Sorted Arrays

20.8%

Hard

下面这个解法真的是太棒了！讲解也讲的很好。基本就是分成两部分，使得右边的始终>=左边的，然后求出左边的max 与 右边的min，然后取中，就可以啦！

真的是太好了！

Link

def median(A, B):
    m, n = len(A), len(B)
    if m > n:
        A, B, m, n = B, A, n, m
    if n == 0:
        raise ValueError

    imin, imax, half_len = 0, m, (m + n + 1) / 2
    while imin <= imax:
        i = (imin + imax) / 2
        j = half_len - i
        if i < m and B[j-1] > A[i]:
            # i is too small, must increase it
            imin = i + 1
        elif i > 0 and A[i-1] > B[j]:
            # i is too big, must decrease it
            imax = i - 1
        else:
            # i is perfect

            if i == 0: max_of_left = B[j-1]
            elif j == 0: max_of_left = A[i-1]
            else: max_of_left = max(A[i-1], B[j-1])

            if (m + n) % 2 == 1:
                return max_of_left

            if i == m: min_of_right = B[j]
            elif j == n: min_of_right = A[i]
            else: min_of_right = min(A[i], B[j])

            return (max_of_left + min_of_right) / 2.0

420

Strong Password Checker

21.0%

Hard

太繁琐，意义不大。

273

Integer to English Words

21.1%

Hard

下面这个代码写的非常的赞。逻辑很清晰。递归和函数都组织的很好。

private final String[] LESS_THAN_20 = {"", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"};
private final String[] TENS = {"", "Ten", "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"};
private final String[] THOUSANDS = {"", "Thousand", "Million", "Billion"};

public String numberToWords(int num) {
    if (num == 0) return "Zero";

    int i = 0;
    String words = "";

    while (num > 0) {
        if (num % 1000 != 0)
            words = helper(num % 1000) +THOUSANDS[i] + " " + words;
        num /= 1000;
        i++;
    }

    return words.trim();
}

private String helper(int num) {
    if (num == 0)
        return "";
    else if (num < 20)
        return LESS_THAN_20[num] + " ";
    else if (num < 100)
        return TENS[num / 10] + " " + helper(num % 10);
    else
        return LESS_THAN_20[num / 100] + " Hundred " + helper(num % 100);
}

30

Substring with Concatenation of All Words

21.7%

Hard

下面代码写的很好。

    // travel all the words combinations to maintain a window
    // there are wl(word len) times travel
    // each time, n/wl words, mostly 2 times travel for each word
    // one left side of the window, the other right side of the window
    // so, time complexity O(wl * 2 * N/wl) = O(2N)
    vector<int> findSubstring(string S, vector<string> &L) {
        vector<int> ans;
        int n = S.size(), cnt = L.size();
        if (n <=  || cnt <= ) return ans;

        // init word occurence
        unordered_map<string, int> dict;
        for (int i = ; i < cnt; ++i) dict[L[i]]++;

        // travel all sub string combinations
        int wl = L[].size();
        for (int i = ; i < wl; ++i) {
            int left = i, count = ;
            unordered_map<string, int> tdict;
            for (int j = i; j <= n - wl; j += wl) {
                string str = S.substr(j, wl);
                // a valid word, accumulate results
                if (dict.count(str)) {
                    tdict[str]++;
                    if (tdict[str] <= dict[str])
                        count++;
                    else {
                        // a more word, advance the window left side possiablly
                        while (tdict[str] > dict[str]) {
                            string str1 = S.substr(left, wl);
                            tdict[str1]--;
                            if (tdict[str1] < dict[str1]) count--;
                            left += wl;
                        }
                    }
                    // come to a result
                    if (count == cnt) {
                        ans.push_back(left);
                        // advance one word
                        tdict[S.substr(left, wl)]--;
                        count--;
                        left += wl;
                    }
                }
                // not a valid word, reset all vars
                else {
                    tdict.clear();
                    count = ;
                    left = j + wl;
                }
            }
        }

        return ans;
    }

440

K-th Smallest in Lexicographical Order

22.1%

Hard

我给出的解法很好（当然了，也是参考了别人的）。使用了前缀、范围等各种技巧来处理。很不错。

140

Word Break II

22.2%

Hard

其实题目不难。用递归，然后用DP加速。开始的时候，我没有一下子想到。开始我也想到用Trie树去查找，但是Trie树会复杂一些，而且不如Hash快。

212

Word Search II

22.2%

Hard

这道题目就是可以用Trie树了。

复习下Trie树的加入。就是26个子树，如果是叶子，标记一下；如果不是叶子，继续递归加。

取的时候，对于每一个格子的元素，如果Trie树找到，就移动到相应的子树，或者是叶子就添加结果；如果Trie树没有找到，就返回。

269

Alien Dictionary

22.3%

Hard

解法很复杂。都要建立图。图是用unordered_map(C++)来处理的。一种使用类似DFS的拓扑排序。注意排序的时候要用visited和path两个数组，前一个是记录是否访问过，后一个是记录是否有环。

另外一种是用BFS的解法，来计算每个节点的入度，然后根据入度来一个一个取出来。

174

Dungeon Game

22.9%

Hard

我之前的方法用的是DP。但是现在我想到了一个很好的方法，就是从右下角开始，往上和往左回溯，来发现最小的代价。

214

Shortest Palindrome

23.0%

Hard

这个在回文类题目系列里面也有写到。通过先反向，再用KMP的next数组来辅助。能够做出来。

另外下面这个方法真是太绝妙了：直接理解有点难。

int j = 0;
    for (int i = s.length() - 1; i >= 0; i--) {
        if (s.charAt(i) == s.charAt(j)) { j += 1; }
    }
    if (j == s.length()) { return s; }
    String suffix = s.substring(j);
    return new StringBuffer(suffix).reverse().toString() + shortestPalindrome(s.substring(0, j)) + suffix;

但是反过来理解，如果中间那段能够是回文，那j必然走到了中间那段的后面，说明中间那段必然不是回文。那么后面的就必然是结果的一部分。再把中间的进行递归，就行了。