Conditions
Questions
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56
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class Solution { public int[][] merge(int[][] intervals) { //sort Arrays.sort(intervals, Comparator.comparingInt(a -> a[0])); List<int[]> merged = new ArrayList<>(); for(int[]interval:intervals){ //empty(),arraylist get last element =merged.size()-1 if( merged.isEmpty() || merged.get(merged.size()-1)[1] < interval[0]){ merged.add(interval); }else if(merged.get(merged.size()-1)[1] >= interval[0] ){ merged.get(merged.size()-1)[1] = Math.max(merged.get(merged.size()-1)[1],interval[1] ); } } //toArray(new int[][]) return merged.toArray(new int[merged.size()][2]); } } - Arrays.sort(intervals, Comparator.comparingInt(a -> a[0]));
Arrays.sort(intervals, (a, b) -> Integer.compare(a[0], b[0]));
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57
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class Solution { public int[][] insert(int[][] intervals, int[] newInterval) { List<int[]> result = new ArrayList<>(); int i = 0; int n = intervals.length; // add all intervals before newInterval while (i < n && intervals[i][1] < newInterval[0]) { result.add(intervals[i]); i++; } // merge overlapping intervals while (i < n && intervals[i][0] <= newInterval[1]) { newInterval[0] = Math.min(newInterval[0], intervals[i][0]); newInterval[1] = Math.max(newInterval[1], intervals[i][1]); i++; } // add merged interval result.add(newInterval); // add remaining intervals while (i < n) { result.add(intervals[i]); i++; } // convert list to array return result.toArray(new int[result.size()][2]); } }
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252
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class Solution { public boolean canAttendMeetings(int[][] intervals) { // Sort the intervals in ascending order of start time Arrays.sort(intervals, (a, b) -> a[0] - b[0]); // Check if any two consecutive meetings overlap for (int i = 0; i < intervals.length - 1; i++) { if (intervals[i][1] > intervals[i+1][0]) { return false; } } // If no overlapping meetings found, return true return true; }
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253
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class Solution { public int minMeetingRooms(int[][] intervals) { // Sort the intervals in ascending order of start time Arrays.sort(intervals, (a, b) -> a[0] - b[0]); // Use a priority queue to keep track of end times PriorityQueue<Integer> endTimes = new PriorityQueue<>(); // Add the end time of the first meeting to the priority queue endTimes.add(intervals[0][1]); // Iterate over the remaining meetings and update the priority queue for (int i = 1; i < intervals.length; i++) { if (intervals[i][0] >= endTimes.peek()) { // If the start time of the current meeting is later than or equal to the earliest ending meeting, // remove the earliest ending meeting from the priority queue and add the end time of the current meeting endTimes.poll(); } endTimes.add(intervals[i][1]); } // Return the size of the priority queue, which represents the minimum number of conference rooms required return endTimes.size(); } }
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435
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class Solution { public int eraseOverlapIntervals(int[][] intervals) { // Sort the intervals in ascending order of end time Arrays.sort(intervals, (a, b) -> a[1] - b[1]); int count = 0; int end = Integer.MIN_VALUE; // Iterate over the intervals and count the number of overlapping intervals for (int[] interval : intervals) { if (interval[0] >= end) { // If the start time of the current interval is later than or equal to the end time of the previous interval, // update the end time and increment the count of non-overlapping intervals end = interval[1]; count++; } } // Return the number of removed intervals, which is the difference between the total number of intervals and the count of non-overlapping intervals return intervals.length - count; } }
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759
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// // Definition for an Interval. // class Interval { // public int start; // public int end; // public Interval() {} // public Interval(int _start, int _end) { // start = _start; // end = _end; // } // }; class Solution { public List<Interval> employeeFreeTime(List<List<Interval>> schedule) { // Use a priority queue to keep track of the next earliest ending interval PriorityQueue<Interval> pq = new PriorityQueue<>((a, b) -> a.start - b.start); // Add all intervals from all employees to the priority queue for (List<Interval> intervals : schedule) { pq.addAll(intervals); } List<Interval> result = new ArrayList<>(); Interval prev = pq.poll(); // Iterate over the intervals in the priority queue and find the free time intervals while (!pq.isEmpty()) { Interval current = pq.poll(); if (prev.end < current.start) { // If there is a gap between the end of the previous interval and the start of the current interval, // add a free time interval to the result result.add(new Interval(prev.end, current.start)); prev = current; } else { // Otherwise, update the previous interval if the end time of the current interval is later if (prev.end < current.end) { prev.end = current.end; } } } return result; } }
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986
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class Solution { public int[][] intervalIntersection(int[][] firstList, int[][] secondList) { List<int[]> result = new ArrayList<>(); int i = 0, j = 0; // Iterate over both lists of intervals and find the intersections while (i < firstList.length && j < secondList.length) { // Find the start and end of the intersection int start = Math.max(firstList[i][0], secondList[j][0]); int end = Math.min(firstList[i][1], secondList[j][1]); if (start <= end) { // If there is an intersection, add it to the result result.add(new int[]{start, end}); } // Move to the next interval with the smaller end time if (firstList[i][1] < secondList[j][1]) { i++; } else { j++; } } // Convert the result to a 2D array and return it return result.toArray(new int[result.size()][]); } }
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1094
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class Solution { public boolean carPooling(int[][] trips, int capacity) { // Create a timeline of events List<int[]> events = new ArrayList<>(); for (int[] trip : trips) { events.add(new int[]{trip[1], trip[0]}); events.add(new int[]{trip[2], -trip[0]}); // Each trip has two events: a pickup event and a dropoff event // The pickup event adds the number of passengers to the car // The dropoff event subtracts the number of passengers from the car } // Sort the events in ascending order of their locations Collections.sort(events, (a, b) -> a[0] - b[0]); // Check the number of passengers at each location int passengers = 0; for (int[] event : events) { passengers += event[1]; if (passengers > capacity) { // If the number of passengers exceeds the car's capacity, return false return false; } } // If the function reaches this point, all trips can be completed return true; } }
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1109
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class Solution { public int[] corpFlightBookings(int[][] bookings, int n) { // Create an array to store the result int[] result = new int[n]; // Iterate through each booking for (int[] booking : bookings) { // Extract the start, end, and seats values from the booking int start = booking[0]; int end = booking[1]; int seats = booking[2]; // Add the seats to the result array for each flight in the booking range for (int i = start; i <= end; i++) { result[i - 1] += seats; } } return result; } }
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495
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class Solution { public int findPoisonedDuration(int[] timeSeries, int duration) { int n = timeSeries.length; if (n == 0) { // if there are no attacks, Ashe is not poisoned return 0; } int total = 0; for (int i = 0; i < n - 1; i++) { // iterate over all attacks except the last one total += Math.min(timeSeries[i+1] - timeSeries[i], duration); // add the duration of poison for each attack } return total + duration; // add duration of poison for the last attack } }
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616
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class Solution { public String addBoldTag(String s, String[] words) { int n = s.length(); boolean[] bold = new boolean[n]; // Mark all positions where words occur in s as bold for (String word : words) { int idx = s.indexOf(word); while (idx != -1) { for (int i = idx; i < idx + word.length(); i++) { bold[i] = true; } idx = s.indexOf(word, idx + 1); } } // Insert bold tags at appropriate positions StringBuilder sb = new StringBuilder(); for (int i = 0; i < n; i++) { if (bold[i] && (i == 0 || !bold[i-1])) { sb.append("<b>"); } sb.append(s.charAt(i)); if (bold[i] && (i == n-1 || !bold[i+1])) { sb.append("</b>"); } } return sb.toString(); } }
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Conclusion
「少种的少收，多种的多收,捐款会鼓励作者多创作以及教会福音传播!!」
少种的少收，多种的多收,捐款会鼓励作者多创作以及教会福音传播!!
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