[LeetCode] 833. Find And Replace in String
grandyang opened this issue · comments
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 <= 1000 < indexes[i] < S.length <= 1000- All characters in given inputs are lowercase letters.
这道题给了我们一个字符串S，并给了一个坐标数组，还有一个源字符串数组，还有目标字符串数组，意思是若某个坐标位置起，源字符串数组中对应位置的字符串出现了，将其替换为目标字符串。题目真的是好长，但好在给了两个例子可以帮助我们很好的理解题意。此题的核心操作就两个，查找和替换，需要注意的是，由于替换操作会改变原字符串，但是我们查找始终是基于最初始的S，比如例子2中，当完成了第一次替换后，S变为了 "eeecd"，好像此时 "ec" 出现了，但仍然不能替换，因为一切查找都是基于最原始的那个S。那么正向的替换可能会产生这样的问题，我们注意到题目中有个限制条件，就是说不会有重叠产生，比如 "abc"，如果让在0位置上查找 "ab" 了，就不会让在1位置上查找 "bc"，这样的话，其实我们可以从后往前开始查找替换，因为不会有重叠，所以后面替换了的字符不会影响到前面。首先我们需要给indexes数组排个序，因为可能不是有序的，但是却不能直接排序，这样会丢失和sources，targets数组的对应关系，这很麻烦。所以我们新建了一个保存pair对儿的数组，将indexes数组中的数字跟其位置坐标组成pair对儿，加入新数组v中，然后给这个新数组按从大到小的方式排序，这样我们既排了序，又保存了对应关系，岂不美哉！
下面就要开始遍历新数组v了，对于遍历到的pair对儿，取出第一个数字，保存到i，表示S中需要查找的位置，取出第二个数字，然后根据这个位置分别到sources和targets数组中取出源字符串和目标字符串，然后我们在S中的i位置，向后取出和源字符串长度相同的子串，然后比较，若正好和源字符串相等，则将其替换为目标字符串即可，参见代码如下：
解法一：
class Solution {
public:
string findReplaceString(string S, vector<int>& indexes, vector<string>& sources, vector<string>& targets) {
vector<pair<int, int>> v;
for (int i = 0; i < indexes.size(); ++i) {
v.push_back({indexes[i], i});
}
sort(v.rbegin(), v.rend());
for (auto a : v) {
int i = a.first;
string s = sources[a.second], t = targets[a.second];
if (S.substr(i, s.size()) == s) {
S = S.substr(0, i) + t + S.substr(i + s.size());
}
}
return S;
}
};
我们也可以使用TreeMap来代替需要排序的数组，由于TreeMap默认的是最小堆，而我们需要的是最大堆，只要在定义上加一个greater就行了，其他部分基本没有任何的区别，参见代码如下：
解法二：
class Solution {
public:
string findReplaceString(string S, vector<int>& indexes, vector<string>& sources, vector<string>& targets) {
map<int, int, greater<int>> m;
for (int i = 0; i < indexes.size(); ++i) {
m[indexes[i]] = i;
}
for (auto a : m) {
int i = a.first;
string s = sources[a.second], t = targets[a.second];
if (S.substr(i, s.size()) == s) {
S = S.substr(0, i) + t + S.substr(i + s.size());
}
}
return S;
}
};
再来看一种稍有不同的解法，之前的两种解法都是直接在S上替换，这里我们新建一个结果res字符串，这里还是使用HashMap来保存映射对，但是稍有不同的是，我们并不是无脑的添加所有的映射对儿，而是先做个check，只要当发现可以查找到源字符串位置的时候，才添加映射对儿，这样就排除了所有不能替换的情况。然后我们遍历原字符串S，对于每个遍历到的位置，我们都到HashMap中查找，如果发现需要替换，我们就把目标字符串提取出来，加入结果res中，注意此时i也需要加上源字符串的长度。若不需要替换，则直接将字符加入结果res中，然后i移动到下一个位置，参见代码如下：
解法三：
class Solution {
public:
string findReplaceString(string S, vector<int>& indexes, vector<string>& sources, vector<string>& targets) {
string res = "";
unordered_map<int, int> m;
for (int i = 0; i < indexes.size(); ++i) {
if (S.substr(indexes[i], sources[i].size()) == sources[i]) {
m[indexes[i]] = i;
}
}
for (int i = 0; i < S.size();) {
if (m.count(i)) {
res += targets[m[i]];
i += sources[m[i]].size();
} else {
res.push_back(S[i]);
++i;
}
}
return res;
}
};
我们也可以使用STL自带的replace函数来完成替换操作，有点偷懒啊～
解法四：
class Solution {
public:
string findReplaceString(string S, vector<int>& indexes, vector<string>& sources, vector<string>& targets) {
map<int, pair<int, string>, greater<int>> m;
for (int i = 0; i < indexes.size(); ++i) {
if (S.substr(indexes[i], sources[i].size()) == sources[i]) {
m[indexes[i]] = {sources[i].size(), targets[i]};
}
}
for (auto a : m) {
S.replace(a.first, a.second.first, a.second.second);
}
return S;
}
};
参考资料：
https://leetcode.com/problems/find-and-replace-in-string/
https://leetcode.com/problems/find-and-replace-in-string/discuss/134758/Java-O(n)-solution