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228. Summary Ranges

228. Summary Ranges

作者: FlynnLWang | 来源:发表于2016-12-27 10:10 被阅读0次

Question

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"].

Code

public class Solution {
    public List<String> summaryRanges(int[] nums) {
        List<String> result = new ArrayList<>();
        if (nums == null || nums.length == 0) return result;
        if (nums.length == 1) {
            result.add(String.valueOf(nums[0]));
            return result;
        }
        
        for (int i = 0; i < nums.length; i++) {
            int a = nums[i];
            while (i + 1 < nums.length && nums[i + 1] - nums[i] == 1) {
                i++;
            }
            if (nums[i] == a) {
                result.add(String.valueOf(a));
            } else {
                result.add(a + "->" + nums[i]);
            }
        }
        return result;
    }
}

Solution

很常规的思路,for循环中不断寻找符合条件的首数字和尾数字即可。

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