821. Shortest Distance to a Character
use std::cmp::min; /** 821. Shortest Distance to a Character https://leetcode.com/problems/shortest-distance-to-a-character/ Given a string s and a character c that occurs in s, return an array of integers answer where answer.length == s.length and answer[i] is the distance from index i to the closest occurrence of character c in s. The distance between two indices i and j is abs(i - j), where abs is the absolute value function. Example 1: Input: s = "loveleetcode", c = "e" Output: [3,2,1,0,1,0,0,1,2,2,1,0] Explanation: The character 'e' appears at indices 3, 5, 6, and 11 (0-indexed). The closest occurrence of 'e' for index 0 is at index 3, so the distance is abs(0 - 3) = 3. The closest occurrence of 'e' for index 1 is at index 3, so the distance is abs(1 - 3) = 2. For index 4, there is a tie between the 'e' at index 3 and the 'e' at index 5, but the distance is still the same: abs(4 - 3) == abs(4 - 5) = 1. The closest occurrence of 'e' for index 8 is at index 6, so the distance is abs(8 - 6) = 2. Example 2: Input: s = "aaab", c = "b" Output: [3,2,1,0] Constraints: 1. 1 <= s.length <= 104 2. s[i] and c are lowercase English letters. 3. It is guaranteed that c occurs at least once in s. */ /* Solution:find out all the index of c and char of s, compare those two index and find out the shortest one; Time:O(n^2), Space:O(n); */ pub struct Solution {} struct Index { c: char, index: i32, } impl Solution { pub fn shortest_to_char(s: String, c: char) -> Vec<i32> { let size = s.len(); let mut indexs: Vec<Index> = Vec::new(); let mut indexs_c: Vec<i32> = Vec::new(); let mut result = Vec::with_capacity(size); for (i,ch) in s.char_indices() { indexs.push(Index { c: ch, index: i as i32 }); if (ch == c) { indexs_c.push(i as i32); } } for item in indexs.iter() { let mut min_value = std::i32::MAX; for i2 in indexs_c.iter() { min_value = std::cmp::min(min_value, (i2 - item.index).abs()); } result.push(min_value); } result } }