【LeetCode】
区间。LeetCode第56题和57题。
56. Insert Interval
下面这种解法是原地插入的方法，显然比新建一个列表省空间。
1 /* Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary). 2 3 You may assume that the intervals were initially sorted according to their start times. 4 5 Example 1: 6 Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9]. 7 8 Example 2: 9 Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16]. 10 11 This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10]. */ 12 13 14 /** 15 * Definition for an interval. 16 * public class Interval { 17 * int start; 18 * int end; 19 * Interval() { start = 0; end = 0; } 20 * Interval(int s, int e) { start = s; end = e; } 21 * } 22 */ 23 class Solution { 24 public List<Interval> insert(List<Interval> intervals, Interval newInterval) { 25 int i = 0; 26 while(i < intervals.size() && intervals.get(i).end < newInterval.start) 27 i++; 28 while(i < intervals.size() && intervals.get(i).start <= newInterval.end){ 29 newInterval = new Interval(Math.min(intervals.get(i).start, newInterval.start), Math.max(intervals.get(i).end, newInterval.end)); 30 intervals.remove(i); 31 } 32 intervals.add(i, newInterval); 33 return intervals; 34 } 35 }
57 Merge Intervals
三种解法，三种思路 ，第一种最快。
1 /* Given a collection of intervals, merge all overlapping intervals. 2 3 For example, 4 Given [1,3],[2,6],[8,10],[15,18], 5 return [1,6],[8,10],[15,18]. */ 6 7 /** 8 * Definition for an interval. 9 * public class Interval { 10 * int start; 11 * int end; 12 * Interval() { start = 0; end = 0; } 13 * Interval(int s, int e) { start = s; end = e; } 14 * } 15 */ 16 17 //Solution1 18 // The idea is that for the result distinct Interval, the latter one's start must > previous one's end. 19 class Solution { 20 public List<Interval> merge(List<Interval> intervals) { 21 int n = intervals.size(); 22 int[] start = new int[n]; 23 int[] end = new int[n]; 24 for(int i = 0; i < n; i++){ 25 start[i] = intervals.get(i).start; 26 end[i] = intervals.get(i).end; 27 } 28 Arrays.sort(start); 29 Arrays.sort(end); 30 List<Interval> res = new ArrayList<>(); 31 for(int i = 0, j = 0; i < n; i++) { 32 if(i == n - 1 || start[i+1] > end[i]){ 33 res.add(new Interval(start[j], end[i])); 34 j = i + 1; 35 } 36 } 37 return res; 38 } 39 } 40 41 //Solution2 42 //The idea is to sort the intervals by their starting points. 43 class Solution { 44 public List<Interval> merge(List<Interval> intervals) { 45 if(intervals.size() <= 1) 46 return intervals; 47 intervals.sort((i1, i2) -> Integer.compare(i1.start, i2.start)); 48 int start = intervals.get(0).start; 49 int end = intervals.get(0).end; 50 List<Interval> res = new ArrayList<>(); 51 for(Interval i : intervals){ 52 if(i.start <= end) 53 end = Math.max(end, i.end); // attention this line, important 54 else{ 55 res.add(new Interval(start, end)); 56 start = i.start; 57 end = i.end; 58 } 59 } 60 res.add(new Interval(start, end)); 61 return res; 62 } 63 } 64 65 //Solution3 66 //use the Insert Interval code, construct a new list, and insert every interval into it 67 //to reuse the existing code or thoughts is very useful and should take care of it
