228 Summary Ranges 汇总区间

给定一个无重复元素的有序整数数组，返回数组中区间范围的汇总。

示例 1:

输入: [0,1,2,4,5,7]
输出: ["0->2","4->5","7"]
示例 2:

输入: [0,2,3,4,6,8,9]
输出: ["0","2->4","6","8->9"]

详见：https://leetcode.com/problems/summary-ranges/description/

Java实现：

class Solution {

    public List<String> summaryRanges(int[] nums) {

        List<String> res=new ArrayList<String>();

        int n=nums.length;

        int i=0;

        while(i<n){

            int j=1;

            while(i+j<n&&nums[i+j]-nums[i]==j){

                ++j;

            }

            res.add(j==1?String.valueOf(nums[i]):String.valueOf(nums[i])+"->"+String.valueOf(nums[i+j-1]));

            i+=j;

        }

        return res;

    }

}

C++实现：

class Solution {

public:

    vector<string> summaryRanges(vector<int>& nums) {

        vector<string> res;

        int i=0,n=nums.size();

        while(i<n)

        {

            int j=1;

            while(i+j<n&&nums[i+j]-nums[i]==j)

            {

                ++j;

            }

            res.push_back(j==1?to_string(nums[i]):to_string(nums[i])+"->"+to_string(nums[i+j-1]));

            i+=j;

        }

        return res;

    }

};

参考：https://www.cnblogs.com/grandyang/p/4603555.html

巴特西

228 Summary Ranges 汇总区间

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