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| 1 | +package weekly; |
| 2 | + |
| 3 | +import java.util.ArrayList; |
| 4 | +import java.util.Arrays; |
| 5 | +import java.util.Collections; |
| 6 | +import java.util.Comparator; |
| 7 | +import java.util.HashMap; |
| 8 | +import java.util.List; |
| 9 | +import java.util.Map; |
| 10 | +import java.util.PriorityQueue; |
| 11 | + |
| 12 | +public class wk302 { |
| 13 | + |
| 14 | + //ranking: 445 / 7092 |
| 15 | + |
| 16 | + //Array |
| 17 | + //简单题,统计每个数字出现多少次就行 |
| 18 | + public int[] numberOfPairs(int[] nums) { |
| 19 | + int[] res = new int[2]; |
| 20 | + int count[] = new int[102]; |
| 21 | + for (int c : nums) { |
| 22 | + count[c]++; |
| 23 | + } |
| 24 | + for (int i : count) { |
| 25 | + res[0] += i / 2; |
| 26 | + res[1] += i % 2; |
| 27 | + } |
| 28 | + return res; |
| 29 | + } |
| 30 | + |
| 31 | + //Heap |
| 32 | + //中等题,将数位和相同的放在一个优先队列中,弹出最大的两个 |
| 33 | + public int maximumSum(int[] nums) { |
| 34 | + Map<Integer, PriorityQueue<Integer>> map = new HashMap<>(); |
| 35 | + for (int num : nums) { |
| 36 | + int c = num; |
| 37 | + int sum = 0; |
| 38 | + while (num > 0) { |
| 39 | + sum += num % 10; |
| 40 | + num /= 10; |
| 41 | + } |
| 42 | + if (!map.containsKey(sum)) { |
| 43 | + map.put(sum, new PriorityQueue<>(Comparator.reverseOrder())); |
| 44 | + } |
| 45 | + map.get(sum).add(c); |
| 46 | + } |
| 47 | + int ans = -1; |
| 48 | + for (Map.Entry<Integer, PriorityQueue<Integer>> entry : map.entrySet()) { |
| 49 | + if (entry.getValue().size() < 2) continue; |
| 50 | + int max = entry.getValue().poll() + entry.getValue().poll(); |
| 51 | + ans = Math.max(ans, max); |
| 52 | + } |
| 53 | + return ans; |
| 54 | + } |
| 55 | + |
| 56 | + class help { |
| 57 | + String value; |
| 58 | + int index; |
| 59 | + |
| 60 | + help(String v, int i) { |
| 61 | + value = v; |
| 62 | + index = i; |
| 63 | + } |
| 64 | + } |
| 65 | + //Heap |
| 66 | + //中等题,先把所有的剪裁结果都保存下来,然后取第k个即可 |
| 67 | + public int[] smallestTrimmedNumbers(String[] nums, int[][] queries) { |
| 68 | + int m = nums[0].length(); |
| 69 | + Map<Integer, List<help>> map = new HashMap<>(); |
| 70 | + for (int i = 0; i < m; i++) { |
| 71 | + List<help> list = new ArrayList<>(); |
| 72 | + for (int j = 0; j < nums.length; j++) { |
| 73 | + String substring = nums[j].substring(i); |
| 74 | + list.add(new help(substring, j)); |
| 75 | + } |
| 76 | + map.put(nums[0].length() - i, list); |
| 77 | + } |
| 78 | + |
| 79 | + for (List<help> value : map.values()) { |
| 80 | + Collections.sort(value, (a, b) -> a.value.compareTo(b.value)); |
| 81 | + } |
| 82 | + |
| 83 | + |
| 84 | + int[] res = new int[queries.length]; |
| 85 | + for (int i = 0; i < queries.length; i++) { |
| 86 | + int[] query = queries[i]; |
| 87 | + List<help> list = map.get(query[1]); |
| 88 | + |
| 89 | + res[i] = list.get(query[0] - 1).index; |
| 90 | + } |
| 91 | + return res; |
| 92 | + } |
| 93 | + |
| 94 | + private static long gcd2(long a, long b) { |
| 95 | + return (a == 0 ? b : gcd2(b % a, a)); |
| 96 | + } |
| 97 | + |
| 98 | + //MATH |
| 99 | + //困难提不困难,求numsDivide的最大公因数,nums从小往大遍历,看哪个能整除即可 |
| 100 | + public int minOperations(int[] nums, int[] numsDivide) { |
| 101 | + Arrays.sort(nums); |
| 102 | + |
| 103 | + int pre = numsDivide[0]; |
| 104 | + for (int i = 0; i < numsDivide.length; i++) { |
| 105 | + pre = (int) gcd2(pre, numsDivide[i]); |
| 106 | + } |
| 107 | + for (int i = 0; i < nums.length; i++) { |
| 108 | + if(pre%nums[i]==0) return i; |
| 109 | + } |
| 110 | + return -1; |
| 111 | + } |
| 112 | + |
| 113 | + public static void main(String[] args) { |
| 114 | + |
| 115 | + } |
| 116 | + |
| 117 | +} |
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