【leetcode】1163. Last Substring in Lexicographical Order

题目如下：

Given a string s, return the last substring of s in lexicographical order.

Example 1:

Input: "abab"
Output: "bab"
Explanation: The substrings are ["a", "ab", "aba", "abab", "b", "ba", "bab"]. The lexicographically maximum substring is "bab".

Example 2:

Input: "leetcode"
Output: "tcode"

Note:

1. 1 <= s.length <= 10^5
2. s contains only lowercase English letters.

解题思路：我的方法是找出s中的最大字符max_char，然后以max_char为分隔符分割s。例如s="azazazzzazbzc"，分割后得到item_list : ['za', 'za', 'zzza', 'zb', 'zc'] ，注意这里舍弃了从字符串头部到第一个max_char之间的部分，同时如果有连续多个max_char出现，合并为同一个子串。接下来遍历item_list,找出最大的子串即可，这里有两点要注意，如果item_list中两个元素相等，那么继续比较这两个元素后面的元素，直到找出不一致为止；另外如果一个item是另一个item的前缀字符串，那么较短的item的值为大。

代码如下：

class Solution(object):
    def lastSubstring(self, s):
        """
        :type s: str
        :rtype: str
        """
        max_char = 'a'
        for i in s:
            max_char = max(max_char,i)

        item_list = []
        sub = ''
        for i in s:
            if sub == '' and i != max_char:
                continue
            elif sub == '' and i == max_char:
                sub += i
            elif sub != '' and i != max_char:
                sub += i
            elif sub != '' and i == max_char and sub[-1] == max_char:
                sub += i
            elif sub != '' and i == max_char and sub[-1] != max_char:
                item_list.append(sub)
                sub = i
            elif sub != '' and i != max_char:
                sub += i
        if len(sub) > 0:item_list.append(sub)

        print item_list

        inx = 0
        sub = item_list[0]
        for i in range(1,len(item_list)):
            if item_list[i] == sub:
                tmp_inx = i + 1
                inx_copy = inx + 1
                while inx_copy < len(item_list) and tmp_inx < len(item_list):
                    if item_list[inx_copy] < item_list[tmp_inx]:
                        sub = item_list[i]
                        inx = i
                        break
                    inx_copy += 1
                    tmp_inx += 1
            elif len(item_list[i]) < len(sub) and item_list[i] == sub[:len(item_list[i])] and i < len(item_list) - 1:
                sub = item_list[i]
                inx = i
            elif sub < item_list[i] and not (len(sub) < len(item_list[i]) and sub == item_list[i][:len(sub)]):
                sub = item_list[i]
                inx = i
        res = ''
        for i in range(inx,len(item_list)):
            res += item_list[i]

        return res