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| 1 | +package weekly; |
| 2 | + |
| 3 | +import java.util.ArrayList; |
| 4 | +import java.util.Arrays; |
| 5 | +import java.util.HashMap; |
| 6 | +import java.util.HashSet; |
| 7 | +import java.util.LinkedList; |
| 8 | +import java.util.List; |
| 9 | +import java.util.Map; |
| 10 | +import java.util.PriorityQueue; |
| 11 | +import java.util.Queue; |
| 12 | +import java.util.Set; |
| 13 | + |
| 14 | +public class wk307 { |
| 15 | + |
| 16 | + //ranking: 659 / 7064 |
| 17 | + |
| 18 | + |
| 19 | + //简单题,注意比赛顺序是从后往前的.. |
| 20 | + public int minNumberOfHours(int initialEnergy, int initialExperience, int[] energy, int[] experience) { |
| 21 | + |
| 22 | + int ans = 0; |
| 23 | + for (int i = 0; i < energy.length; i++) { |
| 24 | + if (energy[i] >= initialEnergy) { |
| 25 | + //严格大于 |
| 26 | + ans += energy[i] - initialEnergy + 1; |
| 27 | + initialEnergy += energy[i] - initialEnergy + 1; |
| 28 | + } |
| 29 | + //精力扣除 |
| 30 | + initialEnergy -= energy[i]; |
| 31 | + |
| 32 | + if (experience[i] >= initialExperience) { |
| 33 | + //严格大于 |
| 34 | + ans += experience[i] - initialExperience + 1; |
| 35 | + initialExperience += experience[i] - initialExperience + 1; |
| 36 | + } |
| 37 | + //经验上涨 |
| 38 | + initialExperience += experience[i]; |
| 39 | + } |
| 40 | + return ans; |
| 41 | + } |
| 42 | + |
| 43 | + |
| 44 | + //中等题,注意边界条件,只需计算前半部分即可 |
| 45 | + public String largestPalindromic(String num) { |
| 46 | + int[] count = new int[10]; |
| 47 | + for (char c : num.toCharArray()) { |
| 48 | + count[c - '0']++; |
| 49 | + } |
| 50 | + StringBuilder sb = new StringBuilder(); |
| 51 | + for (int i = count.length - 1; i >= 0; i--) { |
| 52 | + while (count[i] >= 2) { |
| 53 | + sb.append(i); |
| 54 | + count[i] -= 2; |
| 55 | + } |
| 56 | + } |
| 57 | + boolean add = false; |
| 58 | + int c = -1; |
| 59 | + for (int i = count.length - 1; i >= 0; i--) { |
| 60 | + if (count[i] > 0) { |
| 61 | + add = true; |
| 62 | + c = i; |
| 63 | + break; |
| 64 | + } |
| 65 | + } |
| 66 | + // |
| 67 | + while (sb.length() > 0 && sb.charAt(0) == '0') { |
| 68 | + sb.delete(0, 1); |
| 69 | + } |
| 70 | + //没有单独的>0的数字 |
| 71 | + if (c <= 0 && sb.length() == 0) return "0"; |
| 72 | + StringBuilder reverse = new StringBuilder(sb).reverse(); |
| 73 | + if (!add) { |
| 74 | + return sb.append(reverse).toString(); |
| 75 | + } else { |
| 76 | + return sb.append(c).append(reverse).toString(); |
| 77 | + } |
| 78 | + } |
| 79 | + |
| 80 | + public class TreeNode { |
| 81 | + int val; |
| 82 | + TreeNode left; |
| 83 | + TreeNode right; |
| 84 | + |
| 85 | + TreeNode() { |
| 86 | + } |
| 87 | + |
| 88 | + TreeNode(int val) { |
| 89 | + this.val = val; |
| 90 | + } |
| 91 | + |
| 92 | + TreeNode(int val, TreeNode left, TreeNode right) { |
| 93 | + this.val = val; |
| 94 | + this.left = left; |
| 95 | + this.right = right; |
| 96 | + } |
| 97 | + } |
| 98 | + |
| 99 | + //中等题,先计算图,然后bfs |
| 100 | + public int amountOfTime(TreeNode root, int start) { |
| 101 | + Map<Integer, List<Integer>> map = new HashMap<>(); |
| 102 | + help(root, map, -1); |
| 103 | + Queue<Integer> queue = new LinkedList<>(); |
| 104 | + queue.add(start); |
| 105 | + Set<Integer> set = new HashSet<>(); |
| 106 | + int time = 0; |
| 107 | + set.add(start); |
| 108 | + while (!queue.isEmpty()) { |
| 109 | + int size = queue.size(); |
| 110 | + while (size-- > 0) { |
| 111 | + Integer poll = queue.poll(); |
| 112 | + if (!map.containsKey(poll)) continue; |
| 113 | + for (Integer next : map.get(poll)) { |
| 114 | + if (set.contains(next)) continue; |
| 115 | + set.add(next); |
| 116 | + queue.add(next); |
| 117 | + } |
| 118 | + } |
| 119 | + time++; |
| 120 | + } |
| 121 | + return time - 1; |
| 122 | + } |
| 123 | + |
| 124 | + void help(TreeNode root, Map<Integer, List<Integer>> map, int up) { |
| 125 | + |
| 126 | + if (!map.containsKey(root)) { |
| 127 | + map.put(root.val, new ArrayList<>()); |
| 128 | + } |
| 129 | + if (up != -1) { |
| 130 | + map.get(root.val).add(up); |
| 131 | + } |
| 132 | + if (root.left != null) { |
| 133 | + map.get(root.val).add(root.left.val); |
| 134 | + help(root.left, map, root.val); |
| 135 | + } |
| 136 | + if (root.right != null) { |
| 137 | + map.get(root.val).add(root.right.val); |
| 138 | + help(root.right, map, root.val); |
| 139 | + } |
| 140 | + } |
| 141 | + |
| 142 | + class Pair { |
| 143 | + long sum = 0; |
| 144 | + int index = 0; |
| 145 | + |
| 146 | + Pair(long sum, int index) { |
| 147 | + this.sum = sum; |
| 148 | + this.index = index; |
| 149 | + } |
| 150 | + } |
| 151 | + |
| 152 | + |
| 153 | + //困难题,当时没想出来,看了题解恍然大悟 |
| 154 | + public long kSum(int[] nums, int k) { |
| 155 | + |
| 156 | + //最大的数一定是所有正数的和 |
| 157 | + long sum = 0; |
| 158 | + for (int i = 0; i < nums.length; i++) { |
| 159 | + if (nums[i] > 0) { |
| 160 | + sum += nums[i]; |
| 161 | + } else { |
| 162 | + //变成正数,一律按减法处理 |
| 163 | + nums[i] =-nums[i]; |
| 164 | + } |
| 165 | + } |
| 166 | + if (k == 1) return sum; |
| 167 | + Arrays.sort(nums); |
| 168 | + |
| 169 | + PriorityQueue<Pair> priorityQueue = new PriorityQueue<>((a, b) -> (int) (b.sum - a.sum)); |
| 170 | + priorityQueue.add(new Pair(sum - nums[0], 0)); |
| 171 | + //依次对每个位置取或者不取 |
| 172 | + for(int i=2;i<k;i++){ |
| 173 | + Pair poll = priorityQueue.poll(); |
| 174 | + //System.out.println(poll.sum); |
| 175 | + if (poll.index < nums.length-1) { |
| 176 | + priorityQueue.add(new Pair(poll.sum - nums[poll.index + 1], poll.index + 1)); |
| 177 | + priorityQueue.add(new Pair(poll.sum + nums[poll.index]- nums[poll.index + 1], poll.index + 1)); |
| 178 | + } |
| 179 | + } |
| 180 | + return priorityQueue.poll().sum; |
| 181 | + } |
| 182 | + |
| 183 | + public static void main(String[] args) { |
| 184 | + |
| 185 | + } |
| 186 | +} |
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