[LeetCode] Summary Ranges

Given a sorted integer array without duplicates, return the summary of its ranges.

For example, given [0,1,2,4,5,7], return ["0->2","4->5","7"].

思路:采用双指针  start和end

时间复杂度：O(n)

代码：

    public List<String> summaryRanges(int[] nums) {
        List<String> list=new ArrayList<String>();
        int start=0,end=0;
        while(end++<nums.length)
        {
            if(end==nums.length)
            {
                if(start==end-1)
                    list.add(""+nums[start]);
                else {
                    String string=nums[start]+"->"+nums[end-1];
                    list.add(string);
                }
                break;
            }
            if(nums[end-1]+1==nums[end])
                continue;
            else{
                String string=((start!=end-1)?(nums[start]+"->"+nums[end-1]):""+nums[start]);
                list.add(string);
                start=end;
            }
        }
        return list;
    }

继续优化：

将判断是否最后一个的语句进行整合

    public List<String> summaryRanges(int[] nums) {
        List<String> list=new ArrayList<String>();
        int start=0,end=0;
        while(end++<nums.length)
        {
            if(end!=nums.length && nums[end-1]+1==nums[end])
                continue;
            else{
                String string=((start!=end-1)?(nums[start]+"->"+nums[end-1]):""+nums[start]);
                list.add(string);
                start=end;
            }
        }
        return list;
    }

扩展：假如存在重复的数呢？

    public List<String> summaryRanges(int[] nums) {
        List<String> list=new ArrayList<String>();
        int start=0,end=0;
        while(end++<nums.length)
        {
            if(end!=nums.length && (nums[end-1]+1==nums[end] || nums[end-1]==nums[end]))
                continue;
            else{
                String string=((start!=end-1)?(nums[start]+"->"+nums[end-1]):""+nums[start]);
                list.add(string);
                start=end;
            }
        }
        return list;
    }