From 05003d6c49b4302f63b75f25666fe8c869aa6186 Mon Sep 17 00:00:00 2001 From: markilue <745518019@qq.com> Date: Fri, 5 May 2023 13:22:09 +0800 Subject: [PATCH] =?UTF-8?q?leecode=E6=9B=B4=E6=96=B0?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../difference/LC_1094_CarPooling.java | 25 +++++ .../LC_1109_CorpFlightBookings.java | 106 ++++++++++++++++++ .../second/T38_84_LargestRectangleArea.java | 33 ++++++ .../com/markilue/interview/Question1.java | 56 ++++++++- 4 files changed, 217 insertions(+), 3 deletions(-) create mode 100644 Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/difference/LC_1094_CarPooling.java create mode 100644 Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/difference/LC_1109_CorpFlightBookings.java diff --git a/Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/difference/LC_1094_CarPooling.java b/Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/difference/LC_1094_CarPooling.java new file mode 100644 index 0000000..ccaf8b8 --- /dev/null +++ b/Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/difference/LC_1094_CarPooling.java @@ -0,0 +1,25 @@ +package com.markilue.leecode.hot100.interviewHot.difference; + +/** + *@BelongsProject: Leecode + *@BelongsPackage: com.markilue.leecode.hot100.interviewHot.difference + *@Author: markilue + *@CreateTime: 2023-05-05 11:02 + *@Description: + * TODO 力扣1094 拼车: + * 车上最初有 capacity 个空座位。车 只能 向一个方向行驶（也就是说，不允许掉头或改变方向） + * 给定整数 capacity 和一个数组 trips , trip[i] = [numPassengersi, fromi, toi] 表示第 i 次旅行有 numPassengersi 乘客，接他们和放他们的位置分别是 fromi 和 toi 。 + * 这些位置是从汽车的初始位置向东的公里数。 + * 当且仅当你可以在所有给定的行程中接送所有乘客时，返回 true，否则请返回 false。 + *@Version: 1.0 + */ +public class LC_1094_CarPooling { + + public boolean carPooling(int[][] trips, int capacity) { + + + + return false; + + } +} diff --git a/Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/difference/LC_1109_CorpFlightBookings.java b/Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/difference/LC_1109_CorpFlightBookings.java new file mode 100644 index 0000000..18243de --- /dev/null +++ b/Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/difference/LC_1109_CorpFlightBookings.java @@ -0,0 +1,106 @@ +package com.markilue.leecode.hot100.interviewHot.difference; + +import org.junit.Test; + +import java.util.Arrays; + +/** + *@BelongsProject: Leecode + *@BelongsPackage: com.markilue.leecode.hot100.interviewHot.difference + *@Author: markilue + *@CreateTime: 2023-05-05 11:01 + *@Description: + * TODO 力扣1109 航班预订统计: + * 这里有 n 个航班，它们分别从 1 到 n 进行编号。 + * 有一份航班预订表 bookings ，表中第 i 条预订记录 bookings[i] = [firsti, lasti, seatsi] 意味着在从 firsti 到 lasti （包含 firsti 和 lasti ）的 每个航班 上预订了 seatsi 个座位。 + * 请你返回一个长度为 n 的数组 answer，里面的元素是每个航班预定的座位总数。 + *@Version: 1.0 + */ +public class LC_1109_CorpFlightBookings { + + @Test + public void test(){ + int[][] bookings = {{1, 2, 10}, {2, 3, 20},{2, 5, 25}}; + int n = 5; + System.out.println(Arrays.toString(corpFlightBookings1(bookings,n))); + } + + @Test + public void test1(){ + int[][] bookings = {{1, 2, 10}, {2, 2, 15}}; + int n = 2; + System.out.println(Arrays.toString(corpFlightBookings2(bookings,n))); + } + + + /** + * 本人思路:那么直接构造一个数组把booking挨个加起来 + * 时间复杂度O(n*m) + * 速度击败8.87% 内存击败5.21% 1249ms + * @param bookings + * @param n + * @return + */ + public int[] corpFlightBookings(int[][] bookings, int n) { + + int[] result = new int[n]; + + for (int[] booking : bookings) { + int start = booking[0]; + int end = booking[1]; + while (start <= end) { + result[start++ - 1] += booking[2]; + } + } + + return result; + + } + + + /** + * 差分的思想: + * 当我们希望对原数组的某一个区间 [l,r]施加一个增量inc时，差分数组d对应的改变是：d[l]增加inc，d[r+1]减少inc + * 速度击败100% 内存击败5.21% 2ms + * @param bookings + * @param n + * @return + */ + public int[] corpFlightBookings1(int[][] bookings, int n) { + + int[] result = new int[n+1]; + + for (int[] booking : bookings) { + int start = booking[0]; + int end = booking[1]; + result[start-1] += booking[2]; + result[end] -= booking[2]; + + } + //差分数组最后的结果:10 45 -10 -20 0 -25 + //TODO 为什么result[n+1]对应的位置一定是0? + //因为原本就是0，无论怎么算，后来也是0 + for (int i = 1; i <= n; i++) { + result[i] += result[i - 1]; + } + + return Arrays.copyOf(result,n); + + } + + + public int[] corpFlightBookings2(int[][] bookings, int n) { + + int[] nums = new int[n]; + for (int[] booking : bookings) { + nums[booking[0] - 1] += booking[2]; + if (booking[1] < n) { + nums[booking[1]] -= booking[2]; + } + } + for (int i = 1; i < n; i++) { + nums[i] += nums[i - 1]; + } + return nums; + } +} diff --git a/Leecode/src/main/java/com/markilue/leecode/hot100/second/T38_84_LargestRectangleArea.java b/Leecode/src/main/java/com/markilue/leecode/hot100/second/T38_84_LargestRectangleArea.java index 5689cc7..3435b32 100644 --- a/Leecode/src/main/java/com/markilue/leecode/hot100/second/T38_84_LargestRectangleArea.java +++ b/Leecode/src/main/java/com/markilue/leecode/hot100/second/T38_84_LargestRectangleArea.java @@ -138,4 +138,37 @@ public class T38_84_LargestRectangleArea { return ans; } + + public int largestRectangleArea3(int[] heights) { + int n = heights.length; + int[] left = new int[n]; + int[] right = new int[n]; + + LinkedList stack = new LinkedList<>(); + + for (int i = 0; i < n; i++) { + while (!stack.isEmpty() && heights[stack.peek()] >= heights[i]) {//找到第一个比他小的数就停止 + stack.pop(); + } + left[i] = stack.isEmpty() ? -1 : stack.peek(); + stack.push(i); + } + stack.clear(); + for (int i = n - 1; i >= 0; i--) { + while (!stack.isEmpty() && heights[stack.peek()] >= heights[i]) {//找到第一个比他小的数就停止 + stack.pop(); + } + right[i] = stack.isEmpty() ? n : stack.peek(); + stack.push(i); + } + + int maxResult = 0; + for (int i = 0; i < n; i++) { + maxResult = Math.max((right[i] - left[i] - 1) * heights[i], maxResult); + } + + return maxResult; + + } + } diff --git a/interview/Huawei/src/main/java/com/markilue/interview/Question1.java b/interview/Huawei/src/main/java/com/markilue/interview/Question1.java index 40b9355..83c4d0c 100644 --- a/interview/Huawei/src/main/java/com/markilue/interview/Question1.java +++ b/interview/Huawei/src/main/java/com/markilue/interview/Question1.java @@ -69,9 +69,9 @@ public class Question1 { @Test public void test() { - int[][] nums = new int[][]{{2, 5}, {8, 9}}; -// int[][] nums = new int[][]{{4,8}, {1,6},{2,9}}; - solve1(nums); +// int[][] nums = new int[][]{{2, 5}, {8, 9}}; + int[][] nums = new int[][]{{4, 8}, {1, 6}, {2, 9}}; + solve2(nums); } public void solve(int[][] nums) { @@ -181,4 +181,54 @@ public class Question1 { System.out.println(result); } + + + /** + * 差分法尝试: + * 两个示例都对 + * @param nums + */ + public void solve2(int[][] nums) { + + //1.寻找区间最大值和最小值 + int min = Integer.MAX_VALUE; + int max = Integer.MIN_VALUE; + for (int[] num : nums) { + if (min > num[0]) min = num[0]; + if (max < num[1]) max = num[1]; + } + + //构造差分数组 + int[] diff = new int[max - min + 1]; + + for (int[] num : nums) { + diff[num[0] - min] += 1; + int end = num[1] + 1 - min; + if (end < diff.length) {//超出位置可以不进行计算 + diff[end] -= 1; + } + } + + //通过差分数组还原原本数,并计算这个位置的能耗 + int result = 0; + if (diff[0] == 1) { + result = 3; + } else if (diff[0] >= 2) { + result = 4; + } + for (int i = 1; i < diff.length; i++) { + diff[i] += diff[i - 1]; + if (diff[i] == 1) { + result += 3; + } else if (diff[i] == 0) { + result += 1; + } else { + result += 4; + } + } + + System.out.println(result); + + + } }