LC 833. Find And Replace in String
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.
简单的Median题,用一个index根据index的起始点和长度在原字符串中匹配。注意index可能无序,要先排序以后再用。
Runtime: 4 ms, faster than 99.79% of C++ online submissions for Find And Replace in String.
#define ALL(x) (x).begin(), (x).end()
#define FOR(i, a, b) for (remove_cv<remove_reference<decltype(b)>::type>::type i = (a); i < (b); i++)
#define REP(i, n) FOR(i, 0, n)
#include <vector>
#include <iostream>
#include <unordered_map>
#include <map>
using namespace std;
class Solution {
private:
map<int, vector<string>> mp;
public:
string findReplaceString(string S, vector<int>& indexes, vector<string>& sources, vector<string>& targets) {
REP(i, indexes.size()){
mp[indexes[i]].push_back(sources[i]);
mp[indexes[i]].push_back(targets[i]);
}
int cnt = ;
for(auto it : mp){
indexes[cnt] = it.first;
sources[cnt] = it.second[];
targets[cnt] = it.second[];
cnt++;
}
vector<string> strvec;
string ret;
int idx = ;
REP(i,indexes.size()){
if(idx < indexes[i]){
strvec.push_back(S.substr(idx,indexes[i] - idx));
idx = indexes[i];
}
if(S.substr(indexes[i], sources[i].size()) == sources[i]){
strvec.push_back(targets[i]);
idx += sources[i].size();
}
}
if(idx < S.size()){
strvec.push_back(S.substr(idx));
}
for(auto s : strvec){
ret += s;
}
return ret;
}
};
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