LeetCode 163. Missing Ranges

最新推荐文章于 2024-10-24 09:16:53 发布

原创最新推荐文章于 2024-10-24 09:16:53 发布 · 299 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 题解专栏收录该内容

288 篇文章

订阅专栏

本文介绍了一个寻找有序整数数组中缺失区间的算法实现。给定一个包含特定范围内的整数数组，该算法能够找出所有未出现在数组中的连续整数区间，并以字符串形式返回这些缺失区间。例如，对于输入数组[0,1,3,50,75]及范围[0,99]，输出为[24->4951->7476->99]。

Two pointers.

#include <vector>
#include <string>
#include <iostream>
using namespace std;

/*
  Given a sorted integer array where the range of element 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"];
*/
vector<string> missingRanges(vector<int>& nums, int low, int high) {
  if(nums.size() == 0) return {to_string(low) + "->" + to_string(high)};
  vector<string> res;
  int start = low;
  int end = nums[0];
  for(int i = 0; i < nums.size(); ++i) {
    if(end - start >= 2) {
      string tmp = to_string(start) + "->" + to_string(end - 1);
      res.push_back(tmp);
    }
    start = end + 1;
    end = nums[i];
  }
  if(end - start >= 2) {
    string tmp = to_string(start) + "->" + to_string(end - 1);
    res.push_back(tmp);
  }
  if(high - end >= 2) {
    string tmp = to_string(end + 1) + "->" + to_string(high);
    res.push_back(tmp);
  } else if(high - end >= 1) res.push_back(to_string(high));
  return res;
}

int main(void) {
  vector<int> nums {0, 1, 2, 5, 99};
  vector<string> res = missingRanges(nums, 0, 100);
  for(string s : res) cout << s << endl;
}