seventeen 、 The shortest distance of a character

17.1、 Question design requirements

   Give you a string s And a character c , And c yes s Characters that have appeared in .

   Returns an array of integers answer , among answer.length == s.length And answer[i] yes s From the subscript i To leave it lately The characters of c Of distance .

   Two subscripts i and j Between distance by abs(i - j) , among abs Is an absolute value function .

 Example  1：
 Input ：s = "loveleetcode", c = "e"
 Output ：[3,2,1,0,1,0,0,1,2,2,1,0]
 explain ： character  'e'  Appears in the subscript  3、5、6  and  11  It's about （ Subscript from  0  Start counting ）.
 Distance subscript  0  Current  'e'  Appears in the subscript  3 , So the distance is  abs(0 - 3) = 3 .
 Distance subscript  1  Current  'e'  Appears in the subscript  3 , So the distance is  abs(1 - 3) = 2 .
 For subscripts  4 , Appears in the subscript  3  And subscripts  5  Situated  'e'  Are closest to it , But the distance is the same  abs(4 - 3) == abs(4 - 5) = 1 .
 Distance subscript  8  Current  'e'  Appears in the subscript  6 , So the distance is  abs(8 - 6) = 2 .

 Example  2：
 Input ：s = "aaab", c = "b"
 Output ：[3,2,1,0]

 Tips ：
1 <= s.length <= 104
s[i]  and  c  All are lowercase English letters 
 Topic data assurance  c  stay  s  At least once in 

source ： Power button （LeetCode）
link ：https://leetcode-cn.com/problems/shortest-distance-to-a-character

17.2、 Their thinking

   We need to traverse the string forward to get the result , Then go through it in reverse and get another result , Then compare the two results , take answer Small values can be put together .

17.3、 Algorithm

class Solution {
    
    public int[] shortestToChar(String s, char c) {
    
        int len = s.length();
        int [] answer = new int[len];
        // We need to prev The assignment is divided into two cases , One is a particularly large value , One is a very small value 
        int prev = Integer.MIN_VALUE/2;
        // Positive traversal 
        for (int i = 0; i < len; i++) {
    
            if (s.charAt(i) == c){
    
                prev = i;
            }
            answer[i] = i - prev;
        }
        prev = Integer.MAX_VALUE/2;
        // Reverse traversal 
        for (int i = len-1; i >= 0; i--) {
    
            if (s.charAt(i) == c){
    
                prev = i;
            }
            answer[i] = Math.min(answer[i],prev-i);
        }
        return answer;
    }
}

Reference video ：B standing up The programmer who runs the Marathon