833.Find-And-Replace-in-String

833. Find And Replace in String

题目地址

https://leetcode.com/problems/find-and-replace-in-string/

题目描述

To some string S, we will perform some replacement operations that replace groups of letters with new ones (not necessarily the same size).

Each replacement operation has 3 parameters: a starting index i, a source word x and a target word y.  The rule is that if x starts at position i in the original string S, then we will replace that occurrence of x with y.  If not, we do nothing.

For example, if we have S = "abcd" and we have some replacement operation i = 2, x = "cd", y = "ffff", then because "cd" starts at position 2 in the original string S, we will replace it with "ffff".

Using another example on S = "abcd", if we have both the replacement operation i = 0, x = "ab", y = "eee", as well as another replacement operation i = 2, x = "ec", y = "ffff", this second operation does nothing because in the original string S[2] = 'c', which doesn't match x[0] = 'e'.

All these operations occur simultaneously.  It's guaranteed that there won't be any overlap in replacement: for example, S = "abc", indexes = [0, 1], sources = ["ab","bc"] is not a valid test case.

Example 1:
Input: S = "abcd", indexes = [0,2], sources = ["a","cd"], targets = ["eee","ffff"]
Output: "eeebffff"
Explanation: "a" starts at index 0 in S, so it's replaced by "eee".
"cd" starts at index 2 in S, so it's replaced by "ffff".

Example 2:
Input: S = "abcd", indexes = [0,2], sources = ["ab","ec"], targets = ["eee","ffff"]
Output: "eeecd"
Explanation: "ab" starts at index 0 in S, so it's replaced by "eee". 
"ec" doesn't starts at index 2 in the original S, so we do nothing.

Notes:
0 <= indexes.length = sources.length = targets.length <= 100
0 < indexes[i] < S.length <= 1000
All characters in given inputs are lowercase letters.

代码

Approach #1 Direct

Time: O(NQ) && Space: O(N) where N is the length of S, and we have Q replacement operations.

class Solution {
  public String findReplaceString(String S, int[] indexes, String[] sources, String[] targets) {
        int N = S.length();
    int[] match = new int[N];
    Arrays.fill(match, -1);

    for (int i = 0; i < indexes.length; i++) {
      int ix = indexes[i];
      if (S.substring(ix, ix+sources[i].length()).equals(sources[i])) {
        match[ix] = i;
      }
    }

    StringBuilder ans = new StringBuilder();
    int ix = 0;
    while (ix < N) {
      if (match[ix] >= 0) {
        ans.append(targets[match[ix]]);
        ix += sources[match[ix]].length();
      } else {
        ans.append(S.charAt(ix++));
      }
    }
    return ans.toString();
  }
}