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228. Summary Ranges

时间:2019-11-23 09:30:34      阅读:93      评论:0      收藏:0      [点我收藏+]

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

Example 1:

Input:  [0,1,2,4,5,7]
Output: ["0->2","4->5","7"]
Explanation: 0,1,2 form a continuous range; 4,5 form a continuous range.

Example 2:

Input:  [0,2,3,4,6,8,9]
Output: ["0","2->4","6","8->9"]
Explanation: 2,3,4 form a continuous range; 8,9 form a continuous range.
class Solution {
    public List<String> summaryRanges(int[] nums) {
        int start = 0;
        int end = 0;
        int le = nums.length;
        List<String> res = new ArrayList<String>();
        
        for(int i = 0; i < le; i++){
            start = i;
            while(i + 1< le && nums[i+1] == nums[i] + 1){
                end++;
                i++;
            }
            if(start != end) res.add(nums[start] + "->" + nums[end]);
            if(start == end) res.add(nums[start] + "");
            end++;
        }
        return res;
    }
}

双指针法,碉堡。貌似这种求首尾的题都可以用双指针法。

228. Summary Ranges

原文:https://www.cnblogs.com/wentiliangkaihua/p/11915122.html

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