[LeetCode 641] Missing Ranges

最新推荐文章于 2021-05-09 13:53:21 发布

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本文介绍了LeetCode第641题的解题思路和示例。这道题目要求在给定有序整数数组中找出缺失的范围。通过遍历数组并维护一个变量lastMin来跟踪应有值，当遇到比lastMin大的数时，记录缺失范围。代码实现简洁，能击败22%的提交记录。

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

Example

Example 1

Input:
nums = [0, 1, 3, 50, 75], lower = 0 and upper = 99
Output:
["2", "4->49", "51->74", "76->99"]
Explanation:
in range[0,99],the missing range includes:range[2,2],range[4,49],range[51,74] and range[76,99]

Example 2

Input:
nums = [0, 1, 2, 3, 7], lower = 0 and upper = 7
Output:
["4->6"]
Explanation:
in range[0,7],the missing range include range[4,6]

分析

这道题是一道easy难度的题目，主要就是要注意int的边界的情况。我们使用一个值lastMin记录当前最小的应该值，遍历nums，如果nums[i] == lastMin, 则lastMin = nums[i] + 1；如果nums[i] > lastMin，说明有空缺，就记录下[lastMin, nums[i] -1]。根据lastMin是否等于nums[i] - 1决定最终输出的字符串。

Code

class Solution {
public:
    /*
     * @param nums: a sorted integer array
     * @param lower: An integer
     * @param upper: An integer
     * @return: a list of its missing ranges
     */
    vector<string> findMissingRanges(vector<int> &nums, int lower, int upper) {
        // write your code here
        vector<pair<long long, long long>> ranges;
        long long lastMin = lower;
        for (int i = 0; i < nums.size(); i ++)
        {
            if (nums[i] == lastMin)
            {
                lastMin = nums[i];
                lastMin ++;
            }
            else if (nums[i] > lastMin) 
            {
                ranges.push_back(make_pair(lastMin, nums[i]-1));
                lastMin = nums[i];
                lastMin ++;
            }
        }
        
        if (upper >= lastMin)
        {
            ranges.push_back(make_pair(lastMin, upper));
        }
        
        vector<string> res;
        for (int i = 0; i < ranges.size(); i ++)
        {
            int low = ranges[i].first;
            int up = ranges[i].second;
            if (low == up)
            {
                res.push_back(to_string(low));
            }
            else
            {
                res.push_back(to_string(low) + "->" + to_string(up));
            }
        }
        
        return res;
    }
};

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