【LeetCode】57. Insert Interval

問題描述

https://leetcode.com/problems/insert-interval/#/description

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:
Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].

Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].

This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].

將一個interval插入到已經排序好了的intervals中

算法

因爲已經排序好了，所以只需要一個個遍歷比較看是否需要合併，然後將其添加到結果數組中即可。
遍歷時，設當前interval爲it，會發生以下三種情況：

1. newInterval在it前面，表現爲newInterval.end < it.start，此時先添加newInterval，後添加it，因爲newInterval已經添加進去了，所以後面的就無需比較了
2. newInterval在it後面，表現爲newInterval.start > it.end，此時只添加it到結果數組中即可，讓newInterval再與後面的去比較
3. 兩者有重疊部分，需要合併，將it合併到newInterval中

代碼

public List<Interval> insert(List<Interval> intervals, Interval newInterval) {
            List<Interval> res = new ArrayList<>();
            boolean hasInserted = false; // 是否已插入newInterval
            for(Interval it:intervals) {
                if(hasInserted || it.end < newInterval.start) {
                    res.add(it);
                } else if(it.start > newInterval.end) {
                    res.add(newInterval);
                    res.add(it);
                    hasInserted = true;
                } else {
                    newInterval.start = Math.min(newInterval.start, it.start);
                    newInterval.end = Math.max(newInterval.end, it.end);
                }
            }
            if(!hasInserted) {
                res.add(newInterval);
            }
            return res;
        }

LeetCode解題代碼倉庫：https://github.com/zgljl2012/leetcode-java