830. Positions of Large Groups
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]]
 

    public List<List<Integer>> largeGroupPositions(String S) {
        List<List<Integer>> res = new ArrayList<>();
        if (S == null || S.length() < 3) return res;
        char[] chars = S.toCharArray();
        int i = 0, j = 1 ;
        while (j<=chars.length){
            System.out.println(j);
            if (j ==chars.length) {
                if (j - i >= 3){
                    res.add(new ArrayList<>(Arrays.asList(i,j-1)));
                }
                break;
            }
            if (chars[j] != chars[i]){
                if (j - i >= 3){
                    res.add(new ArrayList<>(Arrays.asList(i,j-1)));
                }
                i = j;
                j++;
            }else {
                j++;
                continue;
            }
        }
        return res;
    }

  public List<List<Integer>> largeGroupPositions(String S) {
        List<List<Integer>> res = new ArrayList<>();
        int left = 0 , right = 1 , count = 1;
        char[] chars = S.toCharArray();
        for (int i = 0; i <chars.length ;i++ ){
           left = i ;
           right = left+1;
            while (right<chars.length && chars[left] == chars[right]){
                right++;
                count++;
            }
            // 上面的循环结束后，right指向一个新的不重复的字符，这里right--,让right指向最后一个重复的字符
            //如果上面的循环没有执行，这里right--了也不要紧，因为在下一轮的循环开始，right会被重新赋值，覆盖原来的值
            right--;
            if (count>=3) {
                res.add(new ArrayList<>(Arrays.asList(left,right)));
                i = right; 
                // 这里将i设为right，在下一个for循环中还有个i++;
                // 让i指向下一个新的字符。这就是为什么不让i= right+1；
            }
            count = 1;
        }
        return res;
    }