weekly contest 327

Author: ZLBer

README.md

@@ -4,7 +4,7 @@ Record leetcode contest and ideas every week, and encourage yourself to think mo
 
 leetcode url: <https://leetcode.cn/u/cctest/>
 
-
+* 🐎️ [weekly contest 327](src/main/java/weekly/wk327.java)     模拟 | 优先队列 | 技术 | 模拟+优先队列
 * 🐎️ [weekly contest 325](src/main/java/weekly/wk325.java)     数学 | 数学 | 动态规划 | 数学
 * 🐎️ [weekly contest 324](src/main/java/weekly/wk324.java)     位运算 | 数学 | 图 | 完全二叉树性质
 * 🐎️ [weekly contest 323](src/main/java/weekly/wk323.java)     排序 | 哈希 | 模拟 | 最小堆、Dijkstra最短路径

src/main/java/weekly/wk327.java

@@ -0,0 +1,164 @@
+package weekly;
+
+import java.util.Comparator;
+import java.util.PriorityQueue;
+
+public class wk327 {
+    //ranking:
+
+    //过
+    public int maximumCount(int[] nums) {
+        int neg = 0, pos = 0;
+        for (int num : nums) {
+            if (num > 0) {
+                pos++;
+            } else if (num < 0) {
+                neg++;
+            }
+        }
+        return Math.max(neg, pos);
+    }
+
+    //模拟
+    public long maxKelements(int[] nums, int k) {
+        PriorityQueue<Integer> priorityQueue = new PriorityQueue<>(Comparator.reverseOrder());
+        for (int num : nums) {
+            priorityQueue.add(num);
+        }
+        long res = 0;
+        while (k-- > 0) {
+            Integer poll = priorityQueue.poll();
+            if (poll == 0) {
+                break;
+            }
+            res += poll;
+            priorityQueue.add((poll + 2) / 3);
+        }
+        return res;
+    }
+
+
+    //统计次数 枚举讨论
+    static public boolean isItPossible(String word1, String word2) {
+        int[] c1 = new int[26];
+        int[] c2 = new int[26];
+        int diff1 = 0, diff2 = 0;
+        for (char c : word1.toCharArray()) {
+            c1[c - 'a']++;
+            if (c1[c - 'a'] == 1) {
+                diff1++;
+            }
+        }
+        for (char c : word2.toCharArray()) {
+            c2[c - 'a']++;
+            if (c2[c - 'a'] == 1) {
+                diff2++;
+            }
+        }
+        //测试交换
+        for (int i = 0; i < c1.length; i++) {
+            for (int j = 0; j < c2.length; j++) {
+                //有一个不存在此字母
+                if (c1[i] == 0 || c2[j] == 0) {
+                    continue;
+                }
+                //相同的字母
+                if (i == j) {
+                    if (diff1 == diff2) {
+                        return true;
+                    } else {
+                        continue;
+                    }
+                }
+                //不同的字母
+                int d1 = diff1, d2 = diff2;
+               //只有一个此字母
+                if (c1[i] == 1) {
+                    d1--;
+                }
+                //没有此字母
+                if (c1[j] == 0) {
+                    d1++;
+                }
+                //只有一个此字母
+                if (c2[j] == 1) {
+                    d2--;
+                }
+                //没有此字母
+                if (c2[i] == 0) {
+                    d2++;
+                }
+                if (d1 == d2) {
+                    return true;
+                }
+            }
+        }
+        return false;
+    }
+
+
+
+    //记住一次一次模拟  不要用while
+    static public int findCrossingTime(int n, int k, int[][] time) {
+        PriorityQueue<int[]> left = new PriorityQueue<>((a, b) -> a[0] + a[2] == b[0] + b[2] ? b[4] - a[4] : b[0] + b[2] - (a[0] + a[2]));
+        PriorityQueue<int[]> right = new PriorityQueue<>((a, b) -> a[0] + a[2] == b[0] + b[2] ? b[4] - a[4] : b[0] + b[2] - (a[0] + a[2]));
+        PriorityQueue<int[]> leftWarehouse = new PriorityQueue<>((a, b) -> a[5] - b[5]);
+        PriorityQueue<int[]> rightWarehouse = new PriorityQueue<>((a, b) -> a[5] - b[5]);
+        for (int i = 0; i < time.length; i++) {
+            int[] t = time[i];
+            left.add(new int[]{t[0], t[1], t[2], t[3], i, 0});
+        }
+
+        int cur = 0;
+        while (n > 0) {
+
+            while (!rightWarehouse.isEmpty() && rightWarehouse.peek()[5] <= cur) {
+                int[] poll = rightWarehouse.poll();
+                right.add(poll);
+            }
+            while (!leftWarehouse.isEmpty() && leftWarehouse.peek()[5] <= cur) {
+                int[] poll = leftWarehouse.poll();
+                left.add(poll);
+            }
+         // 右边的先过河
+            if (!right.isEmpty()) {
+                int[] poll = right.poll();
+                cur += poll[2];
+                poll[5] = cur + poll[3];
+                leftWarehouse.add(poll);
+                //右边空左边可以过
+            } else if (!left.isEmpty()) {
+                int[] poll = left.poll();
+                cur += poll[0];
+                poll[5] = cur + poll[1];
+                rightWarehouse.add(poll);
+                n--;
+              //  右边房子有人
+            } else if (leftWarehouse.isEmpty() && !rightWarehouse.isEmpty()) {
+                cur = rightWarehouse.peek()[5];
+             //左边房子有人
+            } else if (!leftWarehouse.isEmpty() && rightWarehouse.isEmpty()) {
+                cur = leftWarehouse.peek()[5];
+             //都有人取最小
+            } else {
+                cur = Math.min(leftWarehouse.peek()[5], rightWarehouse.peek()[5]);
+            }
+        }
+
+        //将右边的全部过河
+        while (!rightWarehouse.isEmpty()) {
+            int[] poll = rightWarehouse.poll();
+            cur = Math.max(poll[5],cur)+poll[2];
+        }
+
+        return cur;
+
+    }
+
+
+    public static void main(String[] args) {
+        findCrossingTime(3, 2, new int[][]{
+                {1, 9, 1, 8}, {10, 10, 10, 10}
+        });
+    }
+}