|
| 1 | +package weekly; |
| 2 | + |
| 3 | +import java.util.ArrayDeque; |
| 4 | +import java.util.ArrayList; |
| 5 | +import java.util.Arrays; |
| 6 | +import java.util.Deque; |
| 7 | +import java.util.HashMap; |
| 8 | +import java.util.HashSet; |
| 9 | +import java.util.LinkedList; |
| 10 | +import java.util.List; |
| 11 | +import java.util.Map; |
| 12 | +import java.util.Queue; |
| 13 | +import java.util.Set; |
| 14 | + |
| 15 | +public class wk340 { |
| 16 | + |
| 17 | + public static boolean isPrime(int n) { |
| 18 | + if (n == 0 || n == 1) { |
| 19 | + return false; |
| 20 | + } |
| 21 | + for (int i = 2; i <= (int) Math.sqrt(n); i++) { |
| 22 | + if (n % i == 0) { |
| 23 | + return false; |
| 24 | + } |
| 25 | + } |
| 26 | + return true; |
| 27 | + } |
| 28 | + |
| 29 | + //求质数 |
| 30 | + public int diagonalPrime(int[][] nums) { |
| 31 | + int ans = 0; |
| 32 | + for (int i = 0; i < nums.length; i++) { |
| 33 | + if (isPrime(nums[i][i])) { |
| 34 | + ans = Math.max(nums[i][i], ans); |
| 35 | + } |
| 36 | + if (isPrime(nums[i][nums.length - i - 1])) { |
| 37 | + ans = Math.max(nums[i][nums.length - i - 1], ans); |
| 38 | + } |
| 39 | + } |
| 40 | + return ans; |
| 41 | + } |
| 42 | + |
| 43 | + |
| 44 | + // 哈希分组+计算增量或前缀和 |
| 45 | + public long[] distance(int[] nums) { |
| 46 | + long[] arr = new long[nums.length]; |
| 47 | + Map<Integer, List<Integer>> map = new HashMap<>(); |
| 48 | + |
| 49 | + for (int i = 0; i < nums.length; i++) { |
| 50 | + if (!map.containsKey(nums[i])) map.put(nums[i], new ArrayList<>()); |
| 51 | + map.get(nums[i]).add(i); |
| 52 | + } |
| 53 | + for (List<Integer> list : map.values()) { |
| 54 | + long sum = 0; |
| 55 | + for (int i = 1; i < list.size(); i++) { |
| 56 | + sum += list.get(i) - list.get(0); |
| 57 | + } |
| 58 | + |
| 59 | + //最左侧直接求差值和 |
| 60 | + arr[list.get(0)] = sum; |
| 61 | + |
| 62 | + for (int i = 1; i < list.size(); i++) { |
| 63 | + long dis = list.get(i) - list.get(i - 1); |
| 64 | + //增加了i个dis,减少了(list.size-i)个ids |
| 65 | + sum += (i - (list.size() - i)) * dis; |
| 66 | + arr[list.get(i)] = sum; |
| 67 | + } |
| 68 | + |
| 69 | + } |
| 70 | + return arr; |
| 71 | + } |
| 72 | + |
| 73 | + |
| 74 | + /* static public int minimizeMax(int[] nums, int p) { |
| 75 | + PriorityQueue<int[]> priorityQueue = new PriorityQueue<>((a, b) -> a[0] - b[0]); |
| 76 | + Arrays.sort(nums); |
| 77 | + for (int i = 1; i < nums.length; i++) { |
| 78 | + priorityQueue.add(new int[]{nums[i] - nums[i - 1], i - 1, i}); |
| 79 | + } |
| 80 | + Set<Integer> set = new HashSet<>(); |
| 81 | + int ans=0; |
| 82 | + while (p > 0) { |
| 83 | + int[] cur = priorityQueue.poll(); |
| 84 | + if (!set.contains(cur[1]) && !set.contains(cur[2])) { |
| 85 | + set.add(cur[1]); |
| 86 | + set.add(cur[2]); |
| 87 | + ans=Math.max(cur[0],ans); |
| 88 | + int left=cur[1]-1; |
| 89 | + int right=cur[2]+1; |
| 90 | + if(left>=0&&right<nums.length){ |
| 91 | + priorityQueue.add(new int[]{nums[right]-nums[left],left,right}); |
| 92 | + } |
| 93 | + p--; |
| 94 | + } |
| 95 | + } |
| 96 | + return ans; |
| 97 | + }*/ |
| 98 | + |
| 99 | + |
| 100 | + //最大化最小 考虑二分 |
| 101 | + static public int minimizeMax(int[] nums, int p) { |
| 102 | + Arrays.sort(nums); |
| 103 | + int left = 0, right = nums[nums.length - 1] - nums[0]; |
| 104 | + |
| 105 | + while (left < right) { |
| 106 | + int mid = (left + right) / 2; |
| 107 | + if (check(nums, mid, p)) { |
| 108 | + right = mid; |
| 109 | + } else { |
| 110 | + left = mid + 1; |
| 111 | + } |
| 112 | + } |
| 113 | + return left; |
| 114 | + } |
| 115 | + |
| 116 | + static boolean check(int[] nums, int mid, int p) { |
| 117 | + Deque<Integer> deque = new ArrayDeque<>(); |
| 118 | + int ans = 0; |
| 119 | + for (int i = 0; i < nums.length; i++) { |
| 120 | + if (!deque.isEmpty()) { |
| 121 | + if (nums[i] - deque.peekLast() <= mid) { |
| 122 | + ans++; |
| 123 | + deque.pollLast(); |
| 124 | + } else { |
| 125 | + deque.addLast(nums[i]); |
| 126 | + } |
| 127 | + } else { |
| 128 | + deque.addLast(nums[i]); |
| 129 | + } |
| 130 | + } |
| 131 | + return ans >= p; |
| 132 | + } |
| 133 | + |
| 134 | + |
| 135 | + //记录每次每行每列到达的最大位置 |
| 136 | + public int minimumVisitedCells(int[][] grid) { |
| 137 | + Queue<int[]> queue = new LinkedList<>(); |
| 138 | + queue.add(new int[]{0, 0}); |
| 139 | + int m = grid.length; |
| 140 | + int n = grid[0].length; |
| 141 | + int[] maxBelow = new int[n]; |
| 142 | + int[] maxRight = new int[m]; |
| 143 | + |
| 144 | + Set<Integer> set = new HashSet<>(); |
| 145 | + int step = 1; |
| 146 | + set.add(0); |
| 147 | + while (!queue.isEmpty()) { |
| 148 | + int size = queue.size(); |
| 149 | + while (size-- > 0) { |
| 150 | + int[] cur = queue.poll(); |
| 151 | + if (cur[0] == m - 1 && cur[1] == n - 1) { |
| 152 | + return step; |
| 153 | + } |
| 154 | + int k = grid[cur[0]][cur[1]]; |
| 155 | + |
| 156 | + int belowM = Math.min(m - 1, cur[0] + k); |
| 157 | + int rightM = Math.min(n - 1, cur[1] + k); |
| 158 | + |
| 159 | + |
| 160 | + //取最大值,因为可能到不了cur[0]这个位置 |
| 161 | + int begin=Math.max(maxBelow[cur[1]],cur[0]); |
| 162 | + for (int i = begin; i <= belowM; i++) { |
| 163 | + int nx = i, ny = cur[1]; |
| 164 | + if(set.contains(nx * n + ny)) continue; |
| 165 | + set.add(nx * n + ny); |
| 166 | + queue.add(new int[]{nx, ny}); |
| 167 | + } |
| 168 | + //取最大值,因为可能到不了cur[1]这个位置 |
| 169 | + begin=Math.max(maxRight[cur[0]],cur[1]); |
| 170 | + for (int i = begin; i <= rightM; i++) { |
| 171 | + int nx = cur[0], ny = i; |
| 172 | + if(set.contains(nx * n + ny)) continue; |
| 173 | + set.add(nx * n + ny); |
| 174 | + queue.add(new int[]{nx, ny}); |
| 175 | + } |
| 176 | + maxBelow[cur[1]]=Math.max(belowM,maxBelow[cur[1]]); |
| 177 | + maxRight[cur[0]]=Math.max(rightM,maxRight[cur[0]]); |
| 178 | + } |
| 179 | + step++; |
| 180 | + } |
| 181 | + return -1; |
| 182 | + } |
| 183 | + |
| 184 | + public static void main(String[] args) { |
| 185 | + minimizeMax(new int[]{2, 4, 1, 2}, 1); |
| 186 | + } |
| 187 | +} |
0 commit comments