C++实现LeetCode(228.总结区间)

[LeetCode] 228.Summary Ranges 总结区间

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;

									    }

									};