228. Summary Ranges

Posted 2022-01-27 reboot329

刷
June-19-2019

这个题居然也没记过，可能当时觉得太简单了。

楞做法

    public List<String> summaryRanges(int[] nums) 
        List<String> result = new ArrayList<>();
        if (nums.length == 0) return result;
        
        StringBuilder sb = new StringBuilder();
        int prev = nums[0];
        sb.append(prev);
        for (int i = 1; i < nums.length; i ++) 
            int temp = nums[i];
            if (temp == prev + 1) 
                prev = temp;
             else 
                if (prev != Integer.valueOf(sb.toString())) 
                    sb.append("->").append(prev);
                
                result.add(sb.toString());
                sb = new StringBuilder();
                sb.append(temp);
                prev = temp;
            
        
        
        if (prev != Integer.valueOf(sb.toString())) 
                sb.append("->").append(prev);
        
        result.add(sb.toString());
        return result; 
    

然后另一种是通过nums[r] - nums[l] == r - l 来迅速完成l->r
缺点是在[1,3,5,7,9,11]这种每次R都要从最右边到最左边，反而变成O(n2)了

    public List<String> summaryRanges(int[] nums) 
        List<String> result = new ArrayList<>();
        if (nums.length == 0) return result;
        
        int r = nums.length - 1;
        for (int l = 0; l < nums.length; l ++) 
            while (l <= r) 
                if (nums[l] == nums[r]) 
                    result.add(new StringBuilder(Integer.toString(nums[l])).toString());
                    r = nums.length - 1;
                    break;
                 else if (nums[r] - nums[l] == r - l) 
                    result.add(new StringBuilder(Integer.toString(nums[l]))
                                                            .append("->")
                                                            .append(nums[r])
                                                            .toString());
                    l = r;
                    r = nums.length - 1;
                    break;
                 else 
                    r --;
                
            
        

        return result;
    

其实可以二分，类似于归并排序的办法，缺点是merge 时候麻烦点，得处理[1-3][4-5]和[3][4]这种情况= =然后我有点懒就没写。