题目描述：

Given a string S and a character C, return an array of integers representing the shortest distance from the character C in the string.

Example 1:

Input: S = "loveleetcode", C = 'e'Output: [3, 2, 1, 0, 1, 0, 0, 1, 2, 2, 1, 0]

 

Note:

1. S string length is in [1, 10000].
2. C is a single character, and guaranteed to be in string S.
3. All letters in S and C are lowercase.

解题思路：

一直想有没有什么简便的方法，没想出来，就用了笨方法。

代码：

1 class Solution { 2 public: 3     vector
     
       shortestToChar(string S, char C) { 4         vector
      
        res; 5         for (int i = 0; i < S.size(); i++) { 6             if (S[i] == C) 7                 index.push_back(i); 8         } 9         for (int i = 0; i < S.size(); ++i){10             int length = INT_MAX;11             if (S[i] == C) {12                 res.push_back(0);13                 continue;14             }15             for (int j = 0; j < index.size(); ++j) {16                 if (abs(index[j] - i ) < length)17                     length = abs(index[j] - i);18                 else19                     break;20             }21             res.push_back(length);22         }23         return res;24         25     }26     vector
       
         index;27 };
       
      
     

 

---

PS:

 

看了其他人的解法，有一方法很棒，分享下：

1 vector
     
       shortestToChar(const string& s, char c) { 2     int size = s.size(), lastLocation = -1; 3     vector
      
        ret (size, INT_MAX); 4     for (int i = 0; i < size; ++i) { 5         if (s[i] == c) lastLocation = i; 6         if (lastLocation != -1) ret[i] = i - lastLocation; 7     } 8     for (int i = lastLocation - 1; i >= 0; --i) { 9         if (s[i] == c) lastLocation = i;10         ret[i] = min(lastLocation - i, ret[i]);11     }12     return ret;13 }