码迷,mamicode.com
首页 > 其他好文 > 详细

Leetcode-423 Reconstruct Original Digits from English(从英文中重建数字)

时间:2018-08-16 13:55:15      阅读:243      评论:0      收藏:0      [点我收藏+]

标签:class   struct   int   leetcode   tor   character   turn   pre   append   

 1 class Solution
 2 {
 3     public:
 4         string originalDigits(string s)
 5         {
 6             vector<int> CharacterList(26,0);
 7             for(auto c:s)
 8             {
 9                 CharacterList[c-a] ++;
10             }
11             
12             string result;
13             if(CharacterList[g-a])
14             {
15                 result.append(CharacterList[g-a],8);
16                 CharacterList[e-a] -= CharacterList[g-a];
17                 CharacterList[i-a] -= CharacterList[g-a];
18                 CharacterList[h-a] -= CharacterList[g-a];
19                 CharacterList[t-a] -= CharacterList[g-a];
20                 CharacterList[g-a] = 0;
21             }
22             if(CharacterList[x-a])
23             {
24                 result.append(CharacterList[x-a],6);
25                 CharacterList[s-a] -= CharacterList[x-a];
26                 CharacterList[i-a] -= CharacterList[x-a];
27                 CharacterList[x-a] = 0;
28             }
29             if(CharacterList[s-a])
30             {
31                 result.append(CharacterList[s-a],7);
32                 CharacterList[e-a] -= 2*CharacterList[s-a];
33                 CharacterList[v-a] -= CharacterList[s-a];
34                 CharacterList[n-a] -= CharacterList[s-a];
35                 CharacterList[s-a] = 0;
36             }
37             if(CharacterList[v-a])
38             {
39                 result.append(CharacterList[v-a],5);
40                 CharacterList[f-a] -= CharacterList[v-a];
41                 CharacterList[i-a] -= CharacterList[v- a];
42                 CharacterList[e-a] -= CharacterList[v-a];
43                 CharacterList[v-a] = 0;
44             }
45             if(CharacterList[i-a])
46             {
47                 result.append(CharacterList[i-a],9);
48                 CharacterList[n-a] -= 2*CharacterList[i-a];
49                 CharacterList[e-a] -= CharacterList[i-a];
50                 CharacterList[i-a] = 0;
51             }
52             if(CharacterList[u-a])
53             {
54                 result.append(CharacterList[u-a],4);
55                 CharacterList[f-a] -= CharacterList[u-a];
56                 CharacterList[o-a] -= CharacterList[u-a];
57                 CharacterList[r-a] -= CharacterList[u-a];
58                 CharacterList[u-a] = 0;
59             }
60             if(CharacterList[h-a])
61             {
62                 result.append(CharacterList[h-a],3);
63                 CharacterList[t-a] -= CharacterList[h-a];
64                 CharacterList[r-a] -= CharacterList[h-a];
65                 CharacterList[e-a] -= 2*CharacterList[h-a];
66                 CharacterList[h-a] = 0;
67             }
68             if(CharacterList[w-a])
69             {
70                 result.append(CharacterList[w-a],2);
71                 CharacterList[t-a] -= CharacterList[w-a];
72                 CharacterList[o-a] -= CharacterList[w-a];
73                 CharacterList[w-a] = 0;
74             }
75             if(CharacterList[z-a])
76             {
77                 result.append(CharacterList[z-a],0);
78                 CharacterList[e-a] -= CharacterList[z-a];
79                 CharacterList[r-a] -= CharacterList[z-a];
80                 CharacterList[o-a] -= CharacterList[z-a];
81                 CharacterList[z-a] = 0;
82             }
83             if(CharacterList[o-a])
84             {
85                 result.append(CharacterList[o-a],1);
86                 CharacterList[n-a] -= CharacterList[o-a];
87                 CharacterList[e-a] -= CharacterList[o-a];
88                 CharacterList[o-a] = 0;
89             }
90             sort(result.begin(),result.end());
91             return result;
92         }
93 };

 

Leetcode-423 Reconstruct Original Digits from English(从英文中重建数字)

标签:class   struct   int   leetcode   tor   character   turn   pre   append   

原文地址:https://www.cnblogs.com/Asurudo/p/9486368.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!