題目

Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).

You may assume that the intervals were initially sorted according to their start times.

Example 1:

Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
Output: [[1,5],[6,9]]

Example 2:

Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
Output: [[1,2],[3,10],[12,16]]
Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].

思路與解法

intervals列表中的區間interval與newInterval有三種不同情況：①inteval在左，且兩者不相交；②兩者相交；③interval在右，且兩者不相交。
當處於第①種情況時，直接將interval

程式碼實現(GO)

/**
 * Definition for an interval.
 * type Interval struct {
 *     Start int
 *     End   int
 * }
 */
func min(a, b int) int {
    if a < b {
        return a
    }
    return b
}

func max(a, b int) int {
    if
 
 a > b {
        return a
    }
    return b
}


func insert(intervals []Interval, newInterval Interval) []Interval {
    intervals_merged := make([]Interval, 0)
    flag := false
    var i int
    var interval Interval
    // 遍歷整個intervals區間
    for i, interval = range intervals {
        // 當interval右端點 < newInterval左端點，說明兩者無交集，即可將interval加入到intervals_merged中
        if interval.End < newInterval.Start {
            intervals_merged = append(intervals_merged, interval)
        } else if interval.Start > newInterval.End {
            // 當interval左端點 > newInterval右端點，說明兩者無交集，且newInterval不會再次更新，即可將newInterval加入到intervals_merged中
            flag = true
            intervals_merged = append(intervals_merged, newInterval)
            break
        } else {
            // 若interval和newInterval有交集，則更新newInterval
            newInterval.Start = min(interval.Start, newInterval.Start)
            newInterval.End = max(interval.End, newInterval.End)
        }
    }
    // flag == true，說明已經將newInterval加入到intervals_merged中，之後再將intervals[i:]加入即可
    if flag == true {
        intervals_merged = append(intervals_merged, intervals[i:]...)
    } else {
        // flag == false，說明沒有將newInterval加入到intervals_merged，intervals中沒有並未遍歷過的區間
        intervals_merged = append(intervals_merged, newInterval)
    }
    return intervals_merged
}

測試結果