Fork me on GitHub

[LeetCode] 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;
    }
};

 

类似题目:

Missing Ranges

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

 

LeetCode All in One 题目讲解汇总(持续更新中...)

posted @ 2015-06-27 02:25  Grandyang  阅读(10370)  评论(6编辑  收藏
Fork me on GitHub