weekly contest 313

Author: ZLBer

README.md

@@ -30,6 +30,7 @@ leetcode url: <https://leetcode.cn/u/cctest/>
 * 🐶 [weekly contest 290](src/main/java/weekly/wk290.java)
 * 🐭 [weekly contest 280](src/main/java/weekly/wk289.java)
 
+* 🐘 [biweekly contest 87](src/main/java/weekly/wkb88.java)   模拟 | 指针 | 脑筋急转弯 | 线段树
 * 🐘 [biweekly contest 87](src/main/java/weekly/wkb87.java)   模拟 | 双指针 | 位运算 | 贪心
 * 🐘 [biweekly contest 86](src/main/java/weekly/wkb86.java)   数组 | 脑筋急转弯 | 位运算 | 滑动窗口+单调队列
 * 🐹 [biweekly contest 85](src/main/java/weekly/wkb85.java)   滑动窗口 | 模拟 | 差分数组 | 并查集

src/main/java/weekly/wkb88.java

@@ -0,0 +1,162 @@
+package weekly;
+
+import java.util.HashMap;
+import java.util.Map;
+
+public class wkb88 {
+
+    //ranking: 354 / 3905
+    
+    //简单题，暴力解决
+    public boolean equalFrequency(String word) {
+        for (int i = 0; i < word.length(); i++) {
+            int[] count = new int[26];
+            for (int j = 0; j < word.toCharArray().length; j++) {
+                char c = word.charAt(j);
+                if (j == i) continue;
+                count[c - 'a']++;
+            }
+
+            Map<Integer, Integer> map = new HashMap<>();
+            for (int j = 0; j < count.length; j++) {
+                if (count[j] != 0) {
+                    map.put(count[j], map.getOrDefault(count[j], 0) + 1);
+                }
+            }
+            if (map.size() == 1) return true;
+        }
+        return false;
+    }
+
+    //用一个指针标记最长前缀
+    class LUPrefix {
+        int[] dp;
+        int index = 1;
+
+        public LUPrefix(int n) {
+            dp = new int[n + 1];
+            dp[0] = 1;
+        }
+
+        public void upload(int video) {
+            dp[video] = 1;
+            while (index < dp.length && dp[index] == 1) {
+                index++;
+            }
+        }
+
+        public int longest() {
+            return index - 1;
+        }
+    }
+
+    //找规律即可
+    public int xorAllNums(int[] nums1, int[] nums2) {
+        int ans = 0;
+        if (nums2.length % 2 == 1) {
+            for (int i : nums1) {
+                ans ^= i;
+            }
+        }
+        if (nums1.length % 2 == 1) {
+            for (int i : nums2) {
+                ans ^= i;
+            }
+        }
+        return ans;
+    }
+
+    //树状数组板子
+    public class FenwickTree {
+
+        /**
+         * 预处理数组
+         */
+        private int[] tree;
+        private int len;
+
+        public FenwickTree(int n) {
+            this.len = n;
+            tree = new int[n + 1];
+        }
+
+        /**
+         * 单点更新
+         *
+         * @param i     原始数组索引 i
+         * @param delta 变化值 = 更新以后的值 - 原始值
+         */
+        public void update(int i, int delta) {
+            // 从下到上更新，注意，预处理数组，比原始数组的 len 大 1，故 预处理索引的最大值为 len
+            while (i <= len) {
+                tree[i] += delta;
+                i += lowbit(i);
+            }
+        }
+
+        //区间更新
+        void update(int x, int y, int k) {
+            update(x, k);
+            update(y + 1, -k);
+        }
+
+        /**
+         * 查询前缀和
+         *
+         * @param i 前缀的最大索引，即查询区间 [0, i] 的所有元素之和
+         */
+        public int query(int i) {
+            // 从右到左查询
+            int sum = 0;
+            while (i > 0) {
+                sum += tree[i];
+                i -= lowbit(i);
+            }
+            return sum;
+        }
+
+        public int lowbit(int x) {
+            return x & (-x);
+        }
+    }
+
+    //把两个数组拼成一个数组，然后线段树查找指定区间值的个数
+    public long numberOfPairs(int[] nums1, int[] nums2, int diff) {
+        int MAX = 100000;
+        int D = 5 * 10000;
+        for (int i = 0; i < nums1.length; i++) {
+            nums1[i] = nums1[i] - nums2[i];
+        }
+        FenwickTree f = new FenwickTree(MAX);
+        long ans = 0;
+        for (int i = 0; i < nums1.length; i++) {
+            ans += f.query(nums1[i] + diff + D);
+            f.update(nums1[i] + D, 1);
+        }
+        return ans;
+    }
+
+    //补周赛313第四题
+    public int deleteString(String s) {
+        int n = s.length();
+        int[][] lcp = new int[n + 1][n + 1];
+        //求s[i]s[j]的最长公共前缀长度
+        for (int i = n - 1; i >= 0; i--) {
+            for (int j = n - 1; j > i; j--) {
+                if (s.charAt(i) == s.charAt(j)) {
+                    lcp[i][j] = lcp[i + 1][j + 1] + 1;
+                }
+            }
+        }
+        int[] dp = new int[n];
+        for (int i = n-1; i >= 0; i--) {
+            dp[i]=1;
+            for (int j = 1; i + j * 2 <= n; j++) {
+                if (lcp[i][i + j] >= j) {
+                    dp[i]=Math.max(dp[i],dp[i+j]+1);
+                }
+            }
+        }
+        return dp[0];
+    }
+}