思路: 找规律

核心为

$$偶数个数至少为2$$

class Solution {public boolean hasTrailingZeros(int[] nums) {int cnt = 0;for (int v: nums) {if (v% 2 == 0) cnt++;}return cnt >= 2;}
}

和T3一起讲

思路: 同字母分组，然后二分check/top-3分类讨论

利用滑窗，先按同字母进行分组，这样就可以得到连续字母的数字

其实上，拿到某个字母的最大段为m，其答案至少为m-2

这样枚举的话，时间复杂度为$O(n)$

class Solution {public int maximumLength(String s) {char[] str = s.toCharArray();Map<Integer, List<Integer>> hash = new HashMap<>();int i = 0;while (i < str.length) {int p = str[i] - 'a';int j = i + 1;while (j < str.length && str[j] == str[i]) {j++;}int cnt = j - i;hash.computeIfAbsent(p, x -> new ArrayList<>()).add(cnt);i = j;}int res = 0;for (var kv: hash.entrySet()) {List<Integer> lists = kv.getValue();Collections.sort(lists, Comparator.comparing(x -> -x));int l = 0, r = lists.get(0);while (l <= r) {int m = l + (r - l) / 2;int cnt = 0;for (int v: lists) {if (v == m) cnt++;else if (v == m + 1) cnt += 2;else if (v >= m + 2) cnt += 3;}if (cnt >= 3) {l = m + 1;} else {r = m - 1;}}res = Math.max(res, r);}return res == 0 ? -1 : res;}}

思路: 对于回文，其实可以逆转后半段，区间重排是否相等，可以借助前缀和比较字母数即可

因为这题查询量大，而且很难离线处理。

因为这题的思路还是：

$$在线解法$$

那这题的核心就是，如何降低查询的时间复杂度

重排后相等，只要计数相等就行，因此按字母(26)前缀和求解即可

这题还有一个难点，就是改动区间有交集

class Solution {int[] evalute(int[][] pre, int a, int b) {int[] hash = new int[26];for (int i = 0; i < 26; i++) {hash[i] = pre[i][b + 1] - pre[i][a];}return hash;}boolean sub(int[][] pre, int a, int b, int[] hash, int[] hash2) {for (int i = 0; i < 26; i++) {int t = pre[i][b + 1] - pre[i][a];hash[i] -= t;hash2[i] -= t;if (hash[i] < 0) return false;}return true;}public boolean[] canMakePalindromeQueries(String s, int[][] queries) {int n = s.length();int n2 = n / 2;String s1 = s.substring(0, n2);char[] str1 = s1.toCharArray();String s2 = new StringBuilder(s.substring(n2)).reverse().toString();char[] str2 = s2.toCharArray();int[][] pre1 = new int[26][n2 + 1];int[][] pre2 = new int[26][n2 + 1];for (int i = 0; i < 26; i++) {for (int j = 0; j < n2; j++) {pre1[i][j + 1] = pre1[i][j] + (str1[j] - 'a' == i ? 1 : 0);pre2[i][j + 1] = pre2[i][j] + (str2[j] - 'a' == i ? 1 : 0);}}int[] diff = new int[n2 + 1];for (int i = 0; i < n2; i++) {diff[i + 1] = diff[i] + (str1[i] == str2[i] ? 1 : 0);}int m = queries.length;boolean[] res = new boolean[m];for (int i = 0; i < m; i++) {int[] q = queries[i];int a = q[0], b = q[1], c = q[2], d = q[3];int c2 = n - 1 - d;int d2 = n - 1 - c;boolean ok = true;// 分类讨论int lMax = Math.max(a, c2);int rMin = Math.min(b, d2);if (rMin >= lMax) {// 区间有交集int p1 = Math.min(a, c2);int p2 = lMax - 1;int p3 = lMax;int p4 = rMin;int p5 = rMin + 1;int p6 = Math.max(b, d2);if (p1 - 1 >= 0) {if (diff[p1] != p1) {ok = false;}}if (p6 + 1 < n2) {if (diff[n2] - diff[p6 + 1] != n2 - p6 - 1) {ok = false;}}int[] hash1 = evalute(pre1, p1, p6);int[] hash2 = evalute(pre2, p1, p6);// *)if (p2 - p1 >= 0) {if (a == p1) {ok = ok & sub(pre2, p1, p2, hash1, hash2);} else {ok = ok & sub(pre1, p1, p2, hash2, hash1);}}if (p6 - p5 >= 0) {if (b == p6) {ok = ok & sub(pre2, p5, p6, hash1, hash2);} else {ok = ok & sub(pre1, p5, p6, hash2, hash1);}}if (!Arrays.equals(hash1, hash2)) {ok = false;}} else {// 区间没交集int p1 = Math.min(a, c2);int p2 = Math.min(b, d2);int p3 = p2 + 1;int p4 = Math.max(a, c2) - 1;int p5 = Math.max(a, c2);int p6 = Math.max(b, d2);if (p1 - 1 >= 0) {if (diff[p1] != p1) {ok = false;}}if (p6 + 1 < n2) {if (diff[n2] - diff[p6 + 1] != n2 - p6 - 1) {ok = false;}}if (p4 >= p3) {if (diff[p4 + 1] - diff[p3] != p4 - p3 + 1) {ok = false;}}if (p2 >= p1) {int[] hash1 = evalute(pre1, p1, p2);int[] hash2 = evalute(pre2, p1, p2);if (!Arrays.equals(hash1, hash2)) {ok = false;}}if (p6 >= p5) {int[] hash1 = evalute(pre1, p5, p6);int[] hash2 = evalute(pre2, p5, p6);if (!Arrays.equals(hash1, hash2)) {ok = false;}}}res[i] = ok;}return res;}}