[LeetCode] 228.Summary Ranges 總結(jié)區(qū)間

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.

這道題給定我們一個有序數(shù)組，讓我們總結(jié)區(qū)間，具體來說就是讓我們找出連續(xù)的序列，然后首尾兩個數(shù)字之間用個“->"來連接，那么我只需遍歷一遍數(shù)組即可，每次檢查下一個數(shù)是不是遞增的，如果是，則繼續(xù)往下遍歷，如果不是了，我們還要判斷此時是一個數(shù)還是一個序列，一個數(shù)直接存入結(jié)果，序列的話要存入首尾數(shù)字和箭頭“->"。我們需要兩個變量i和j，其中i是連續(xù)序列起始數(shù)字的位置，j是連續(xù)數(shù)列的長度，當j為1時，說明只有一個數(shù)字，若大于1，則是一個連續(xù)序列，代碼如下：

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;
    }
};

類似題目：

Missing Ranges

Data Stream as Disjoint Intervals 

參考資料：

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

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