[LeetCode] Missing Ranges

最新推荐文章于 2021-05-09 13:53:21 发布

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最新推荐文章于 2021-05-09 13:53:21 发布

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原文链接：http://www.cnblogs.com/jcliBlogger/p/4601221.html

本文深入探讨了如何通过高效算法确定已排序整数数组中缺失的区间范围，详细介绍了从给定数组和范围边界出发，通过迭代计算并返回缺失区间列表的过程。文章以代码实现为例，展示了如何统一处理不同情况，确保代码简洁且易于理解。

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Problem Description:

Given a sorted integer array where the range of elements are [lower, upper] inclusive, return its missing ranges.

For example, given [0, 1, 3, 50, 75], lower = 0 and upper = 99, return ["2", "4->49", "51->74", "76->99"].

Well, I guess this problem aims at testing your ability to think throughly. The tricky part in this problem is to take all possible cases into consideration and unify them to give a succinct code.

The code is as follows.

 1 class Solution {
 2 public:
 3     vector<string> findMissingRanges(vector<int>& nums, int lower, int upper) {
 4         vector<string> ranges;
 5         int pre = lower - 1, n = nums.size();
 6         for (int i = 0; i <= n; i++) {
 7             int cur = (i == n) ? upper + 1 : nums[i];
 8             if (cur - pre >= 2)
 9                 ranges.push_back(missingRanges(pre + 1, cur - 1));
10             pre = cur;
11         }
12         return ranges;
13     }
14 private:
15     string missingRanges(int low, int up) {
16         return (low == up) ? to_string(low) : to_string(low) + "->" + to_string(up);
17     }
18 };