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.
Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.
这道题给定我们一个有序数组，让我们总结区间，具体来说就是让我们找出连续的序列，然后首尾两个数字之间用个“->"来连接，那么我只需遍历一遍数组即可，每次检查下一个数是不是递增的，如果是，则继续往下遍历，如果不是了，我们还要判断此时是一个数还是一个序列，一个数直接存入结果，序列的话要存入首尾数字和箭头“->"。我们需要两个变量i和j，其中i是连续序列起始数字的位置，j是连续数列的长度，当j为1时，说明只有一个数字，若大于1，则是一个连续序列，代码如下：
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 && (long)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; } };
类似题目：
Data Stream as Disjoint Intervals
参考资料：
https://leetcode.com/problems/summary-ranges/
https://leetcode.com/problems/summary-ranges/discuss/63451/9-lines-c%2B%2B-0ms-solution
https://leetcode.com/problems/summary-ranges/discuss/63219/Accepted-JAVA-solution-easy-to-understand