[LeetCode] Positions of Large Groups 大羣組的位置

 

In a string S of lowercase letters, these letters form consecutive groups of the same character.

For example, a string like S = "abbxxxxzyy" has the groups "a", "bb", "xxxx", "z" and "yy".

Call a group large if it has 3 or more characters.  We would like the starting and ending positions of every large group.

The final answer should be in lexicographic order.

 

Example 1:

Input: "abbxxxxzzy"
Output: [[3,6]]
Explanation: large group with starting  3 and ending positions 6.
"xxxx" is the single

Example 2:

Input: "abc"
Output: []
Explanation: We have "a","b" and "c" but no large group.

Example 3:

Input: "abcdddeeeeaabbbcd"
Output: [[3,5],[6,9],[12,14]]

 

Note:  1 <= S.length <= 1000

 

這道題給了咱們一個全小寫的字符串，說是重複出現的字符能夠看成一個羣組，若是重複次數大於等於3次，能夠看成一個大羣組，讓咱們找出全部大羣組的起始和結束位置。那麼實際上就是讓咱們計數連續重複字符的出現次數，因爲要連續，因此咱們能夠使用雙指針來作，一個指針指向重複部分的開頭，一個日後遍歷計數，只要不相同了就中止，而後看次數是否大於等3，是的話就將雙指針位置存入結果res中，並更新指針，參見代碼以下：

 

解法一：

class Solution {
public:
    vector<vector<int>> largeGroupPositions(string S) {
        vector<vector<int>> res;
        int n = S.size(), i = 0, j = 0;
        while (j < n) {
            while (j < n && S[j] == S[i]) ++j;
            if (j - i >= 3) res.push_back({i, j - 1});
            i = j;
        }
        return res;
    }
};

 

咱們也能夠換一種寫法，不用while循環，而是使用for循環，但本質上仍是雙指針的思路，並無什麼太大的區別，參見代碼以下：

 

解法二：

class Solution {
public:
    vector<vector<int>> largeGroupPositions(string S) {
        vector<vector<int>> res;
        int n = S.size(), start = 0;
        for (int i = 1; i <= n; ++i) {
            if (i < n && S[i] == S[start]) continue;
            if (i - start >= 3) res.push_back({start, i - 1});
            start = i;
        }
        return res;
    }
};

 

參考資料：

https://leetcode.com/problems/positions-of-large-groups/

https://leetcode.com/problems/positions-of-large-groups/discuss/128961/Java-Solution-Two-Pointers

https://leetcode.com/problems/positions-of-large-groups/discuss/128942/My-Easy-7-Lines-C%2B%2B-Solution

 

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