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: "xxxx" is the single large group with starting 3 and ending positions 6.
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
这道题给了我们一个全小写的字符串，说是重复出现的字符可以当作一个群组，如果重复次数大于等于3次，可以当作一个大群组，让我们找出所有大群组的起始和结束位置。那么实际上就是让我们计数连续重复字符的出现次数，由于要连续，所以我们可以使用双指针来做，一个指针指向重复部分的开头，一个往后遍历计数，只要不相同了就停止，然后看次数是否大于等3，是的话就将双指针位置存入结果res中，并更新指针，参见代码如下：
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; } };
我们也可以换一种写法，不用while循环，而是使用for循环，但本质上还是双指针的思路，并没有什么太大的区别，参见代码如下：
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; } };
参考资料：
https://leetcode.com/problems/positions-of-large-groups/
https://leetcode.com/problems/positions-of-large-groups/discuss/128961/Java-Solution-Two-Pointers