LeetCode-821. Shortest Distance to a Character

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.

思路：

1.首先看常规做法，定义vector<int>ans(n,n),n=S.size(),存放答案 ,遍历字符串,对每个字符,如果是C则令ans[i]=0,否则使用while()循环分别向左,右寻找C,然后取较小者.Time Complexity O(N^2).

参考代码:

class Solution {
public:
    bool ht[10010]={false};
    vector<int> shortestToChar(string S, char C) {
        vector<int> ans(S.size(),S.size());
        for(int i=0;i<S.size();i++){
            if(S[i]==C) ans[i]=0;
            else{
                int l=i,r=i;
                while(l>=0&&S[l]!=C) l--;
                while(r<S.size()&&S[r]!=C) r++;
                if(l>=0&&S[l]==C&&r<S.size()&&S[r]==C) ans[i]=min(i-l,r-i);
                else if(l>=0&&S[l]==C) ans[i]=i-l;
                else if(r<S.size()&&S[r]==C) ans[i]=r-i;
            }
        }
        return ans;
    }
};

2.优化：首先遍历一遍,若S[i]==C,令ans[i]=0 .然后分别从左边和右边进行迭代求最小距离.

参考代码:

class Solution {
public:
    bool ht[10010]={false};
    vector<int> shortestToChar(string S, char C) {
        int n=S.size();
        vector<int> ans(n,n);
        for(int i=0;i<n;i++)   if(S[i]==C) ans[i]=0;
        for(int i=1;i<n;i++) ans[i]=min(ans[i],ans[i-1]+1);
        for(int i=n-2;i>=0;--i) ans[i]=min(ans[i],ans[i+1]+1);
        return ans;
    }
};