821. Shortest Distance to a Character

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：

class Solution {
public:
    vector<int> shortestToChar(string S, char C) {
        vector<int> res;
        for (int i = 0; i < S.size(); i++) {
            if (S[i] == C)
                index.push_back(i);
        }
        for (int i = 0; i < S.size(); ++i){
            int length = INT_MAX;
            if (S[i] == C) {
                res.push_back(0);
                continue;
            }
            for (int j = 0; j < index.size(); ++j) {
                if (abs(index[j] - i ) < length)
                    length = abs(index[j] - i);
                else
                    break;
            }
            res.push_back(length);
        }
        return res;
        
    }
    vector<int> index;
};

 

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PS:

看了其他人的解法，有一方法很棒，分享下：

vector<int> shortestToChar(const string& s, char c) {
    int size = s.size(), lastLocation = -1;
    vector<int> ret (size, INT_MAX);
    for (int i = 0; i < size; ++i) {
        if (s[i] == c) lastLocation = i;
        if (lastLocation != -1) ret[i] = i - lastLocation;
    }
    for (int i = lastLocation - 1; i >= 0; --i) {
        if (s[i] == c) lastLocation = i;
        ret[i] = min(lastLocation - i, ret[i]);
    }
    return ret;
}