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题单来源： https://leetcode.cn/problems/minimum-size-subarray-in-infinite-array/solutions/2464878/hua-dong-chuang-kou-on-shi-jian-o1-kong-cqawc/

3. 无重复字符的最长子串

https://leetcode.cn/problems/longest-substring-without-repeating-characters/description/

提示：
0 <= s.length <= 5 * 10^4
s 由英文字母、数字、符号和空格组成

枚举右端点，根据窗口内计数情况移动左端点。

class Solution {public int lengthOfLongestSubstring(String s) {int ans = 0;int[] cnt = new int[128];for (int l = 0, r = 0; r < s.length(); ++r) {char ch = s.charAt(r);cnt[ch]++;while (cnt[ch] > 1) cnt[s.charAt(l++)]--;ans = Math.max(r - l + 1, ans);}return ans;}
}

1493. 删掉一个元素以后全为 1 的最长子数组

https://leetcode.cn/problems/longest-subarray-of-1s-after-deleting-one-element/description/

提示：

1 <= nums.length <= 10^5
nums[i] 要么是 0 要么是 1 。

窗口内最多有 1 个0，用 id 记录上一个出现 0 的位置，当出现新的 0 时，将左端点设置为 l = id + 1。 这样就保持了窗口内始终最多有 1 个 0。

class Solution {public int longestSubarray(int[] nums) {int n = nums.length, ans = 0;for (int l = 0, r = 0, id = -1; r < n; ++r) {if (nums[r] == 0) {l = id + 1;id = r;}ans = Math.max(r - l, ans);}return ans;}
}

904. 水果成篮

https://leetcode.cn/problems/fruit-into-baskets/description/

提示：

1 <= fruits.length <= 10^5
0 <= fruits[i] < fruits.length

class Solution {public int totalFruit(int[] fruits) {Map<Integer, Integer> m = new HashMap<>();int ans = 0;for (int l = 0, r = 0; r < fruits.length; ++r) {m.merge(fruits[r], 1, Integer::sum);while (m.size() > 2) {m.merge(fruits[l], -1, Integer::sum);if (m.get(fruits[l]) == 0) m.remove(fruits[l]);l++;}ans = Math.max(ans, r - l + 1);}return ans;}
}

1695. 删除子数组的最大得分

https://leetcode.cn/problems/maximum-erasure-value/description/

提示：

1 <= nums.length <= 10^5
1 <= nums[i] <= 10^4

class Solution {public int maximumUniqueSubarray(int[] nums) {int n = nums.length, ans = 0, s = 0;int[] cnt = new int[10001];for (int l = 0, r = 0; r < n; ++r) {s += nums[r];cnt[nums[r]]++;while (cnt[nums[r]] > 1) {s -= nums[l];cnt[nums[l++]]--;}if (s > ans) ans = s;}return ans;}
}

用 s > ans 判断会比使用 s = Math.max(ans, s) 快一些。

2841. 几乎唯一子数组的最大和

https://leetcode.cn/problems/maximum-sum-of-almost-unique-subarray/description/

提示：

1 <= nums.length <= 2 * 10^4
1 <= m <= k <= nums.length
1 <= nums[i] <= 10^9

class Solution {public long maxSum(List<Integer> nums, int m, int k) {long ans = 0, n = nums.size(), s = 0;Map<Integer, Integer> cnt = new HashMap<>();for (int i = 0; i < k - 1; ++i) {s += nums.get(i);cnt.merge(nums.get(i), 1, Integer::sum);}for (int l = 0, r = k - 1; r < n; ++l, ++r) {s += nums.get(r);cnt.merge(nums.get(r), 1, Integer::sum);if (cnt.size() >= m) ans = Math.max(ans, s);s -= nums.get(l);cnt.merge(nums.get(l), -1, Integer::sum);if (cnt.get(nums.get(l)) == 0) cnt.remove(nums.get(l));}return ans;}
}

2024. 考试的最大困扰度

https://leetcode.cn/problems/maximize-the-confusion-of-an-exam/description/

提示：
n == answerKey.length
1 <= n <= 5 * 10^4
answerKey[i] 要么是 'T' ，要么是 'F'
1 <= k <= n

维护一个滑动窗口，枚举右端点，当窗口中T和F的较小数量大于k时，将左端点向右移。
在这个过程中用窗口长度更新答案。

class Solution {public int maxConsecutiveAnswers(String answerKey, int k) {int n = answerKey.length(), ans = 0;int t = 0, f = 0;for (int l = 0, r = 0; r < n; ++r) {char a = answerKey.charAt(r);if (a == 'T') t++;else f++;// 将左端点向右移while (t > k && f > k) {a = answerKey.charAt(l++);if (a == 'T') t--;else f--;}ans = Math.max(r - l + 1, ans);}return ans;}
}

1004. 最大连续1的个数 III

https://leetcode.cn/problems/max-consecutive-ones-iii/description/

提示：

1 <= nums.length <= 10^5
nums[i] 不是 0 就是 1
0 <= k <= nums.length

维护一个窗口中最多出现 k 个 0 的滑动窗口。

class Solution {public int longestOnes(int[] nums, int k) {int n = nums.length, c0 = 0, ans = 0;for (int l = 0, r = 0; r < n; ++r) {if (nums[r] == 0) c0++;while (c0 > k) {if (nums[l++] == 0) c0--;}ans = Math.max(r - l + 1, ans);}return ans;}
}

1438. 绝对差不超过限制的最长连续子数组

https://leetcode.cn/problems/longest-continuous-subarray-with-absolute-diff-less-than-or-equal-to-limit/description/

提示：

1 <= nums.length <= 10^5
1 <= nums[i] <= 10^9
0 <= limit <= 10^9

同时维护窗口内的最大值和最小值。
可以使用 TreeMap 或者 单调队列来做。

class Solution {public int longestSubarray(int[] nums, int limit) {TreeMap<Integer, Integer> tm = new TreeMap<>();int ans = 0;for (int l = 0, r = 0; r < nums.length; ++r) {tm.merge(nums[r], 1, Integer::sum);while (tm.lastKey() - tm.firstKey() > limit) {tm.merge(nums[l], -1, Integer::sum);if (tm.get(nums[l]) == 0) tm.remove(nums[l]);l++;}ans = Math.max(r - l + 1, ans);}return ans;}
}

2401. 最长优雅子数组

https://leetcode.cn/problems/longest-nice-subarray/description/

提示：
1 <= nums.length <= 10^5
1 <= nums[i] <= 10^9

解法1——维护窗口内int各位出现的次数

题目要求即为int的32位中，每一位都最多出现一次。

class Solution {public int longestNiceSubarray(int[] nums) {int n = nums.length, ans = 0;int[] cnt = new int[32];for (int l = 0, r = 0; r < n; ++r) {op(cnt, nums[r], 1);while (!check(cnt)) op(cnt, nums[l++], -1);ans = Math.max(ans, r - l + 1);}return ans;}public void op(int[] cnt, int x, int m) {for (int i = 0; i < cnt.length; ++i) {if ((x >> i & 1) == 1) cnt[i] += m;}}public boolean check(int[] cnt) {for (int c: cnt) {if (c > 1) return false;}return true;}
}

解法2——利用位运算的性质 维护窗口🐂

充分 利用了 & ^ | 三种运算的性质。

使用 & 判断是否有交集。
使用 ^ 模拟去除。
使用 | 模拟加入。

class Solution {public int longestNiceSubarray(int[] nums) {int ans = 0;for (int left = 0, right = 0, or = 0; right < nums.length; right++) {// 有交集while ((or & nums[right]) > 0) {or ^= nums[left++]; // 从 or 中去掉集合 nums[left]}or |= nums[right];      // 把集合 nums[right] 并入 or 中ans = Math.max(ans, right - left + 1);}return ans;}
}

1658. 将 x 减到 0 的最小操作数

https://leetcode.cn/problems/minimum-operations-to-reduce-x-to-zero/description/

提示：
1 <= nums.length <= 10^5
1 <= nums[i] <= 10^4
1 <= x <= 10^9

等价于找到窗口内和为 sum - x 的最长子数组的长度。

class Solution {public int minOperations(int[] nums, int x) {// 等价于找到窗口内和为 sum - x 的最长子数组的长度。int n = nums.length, t = Arrays.stream(nums).sum() - x, ans = -1, s = 0;for (int l = 0, r = 0; r < n; ++r) {s += nums[r];while (l <= r && s > t) s -= nums[l++];if (s == t) ans = Math.max(ans, r - l + 1); }return ans != -1? n - ans: -1;}
}

1838. 最高频元素的频数

https://leetcode.cn/problems/frequency-of-the-most-frequent-element/description/

提示：
1 <= nums.length <= 10^5
1 <= nums[i] <= 10^5
1 <= k <= 10^5

排序 + 滑动窗口。
每个窗口都假设变成最大的那个数字，不能变的移出窗口。

class Solution {public int maxFrequency(int[] nums, int k) {Arrays.sort(nums);int n = nums.length, ans = 1;long s = 0;for (int l = 0, r = 0; r < n; ++r) {s += nums[r];while ((long)(r - l + 1) * nums[r] - k > s) s -= nums[l++];if ((long)(r - l + 1) * nums[r] - k <= s) ans = Math.max(ans, r - l + 1);}return ans;}
}

2831. 找出最长等值子数组

https://leetcode.cn/problems/find-the-longest-equal-subarray/description/

提示：
1 <= nums.length <= 10^5
1 <= nums[i] <= nums.length
0 <= k <= nums.length

解法1——双哈希表：频次哈希表 和 频次的频次哈希表

频次的频次哈希表 用于 快速找到最大的频次。

class Solution {public int longestEqualSubarray(List<Integer> nums, int k) {int n = nums.size(), ans = 0;Map<Integer, Integer> cnt = new HashMap<>();// 频次 哈希表 ，用于快速找到最大的频次TreeMap<Integer, Integer> pCnt = new TreeMap<>();for (int l = 0, r = 0; r < n; ++r) {cnt.merge(nums.get(r), 1, Integer::sum);pCnt.merge(cnt.get(nums.get(r)), 1, Integer::sum);pCnt.merge(cnt.get(nums.get(r)) - 1, -1, Integer::sum);while (r - l + 1 - pCnt.lastKey() > k) {cnt.merge(nums.get(l), -1, Integer::sum);pCnt.merge(cnt.get(nums.get(l)), 1, Integer::sum);pCnt.merge(cnt.get(nums.get(l)) + 1, -1, Integer::sum);if (pCnt.get(cnt.get(nums.get(l)) + 1) == 0) pCnt.remove(cnt.get(nums.get(l)) + 1);l++;}ans = Math.max(pCnt.lastKey(), ans);}return ans;}
}

解法2——分组 + 双指针 🐂

https://leetcode.cn/problems/find-the-longest-equal-subarray/solutions/2396401/fen-zu-shuang-zhi-zhen-pythonjavacgo-by-lqqau/

算法思路见下图。

class Solution {public int longestEqualSubarray(List<Integer> nums, int k) {int n = nums.size(), ans = 0;Map<Integer, List<Integer>> m = new HashMap<>();for (int i = 0; i < n; ++i) {if (!m.containsKey(nums.get(i))) {m.put(nums.get(i), new ArrayList<>());}List<Integer> ls = m.get(nums.get(i));ls.add(i - ls.size());}for (List<Integer> ls: m.values()) {int res = 0;for (int l = 0, r = 0; r < ls.size(); ++r) {while (ls.get(r) - ls.get(l) > k) l++;res = Math.max(res, r - l + 1);}ans = Math.max(ans, res);}return ans;}
}

2106. 摘水果

https://leetcode.cn/problems/maximum-fruits-harvested-after-at-most-k-steps/description/

维护一个窗口，窗口中是可以被采集的水果。
随着右端点的枚举，左端点也会随着窗口的扩大导致不合理，从而将左端点右移。

class Solution {public int maxTotalFruits(int[][] fruits, int startPos, int k) {int ans = 0, s = 0;for (int l = 0, r = 0; r < fruits.length; ++r) {s += fruits[r][1];while (l <= r && !check(fruits[l][0], fruits[r][0], startPos, k)) s-= fruits[l++][1];ans = Math.max(ans, s);}return ans;}// 检查窗口是否合理public boolean check(int a, int b, int x, int k) {if (a <= x && b <= x) return x - a <= k;    // 都在左边if (a >= x && b >= x) return b - x <= k;    // 都在右边// a在左边，b在右边a = x - a;b = b - x;return Math.min(a, b) * 2 + Math.max(a, b) <= k;}
}

1610. 可见点的最大数目⭐（坐标转换成极角）

https://leetcode.cn/problems/maximum-number-of-visible-points/description/

提示：
1 <= points.length <= 10^5
points[i].length == 2
location.length == 2
0 <= angle < 360
0 <= posx, posy, xi, yi <= 100

坐标转成极角的方法为 Math.atan2()，见：https://docs.oracle.com/en/java/javase/11/docs/api/java.base/java/lang/Math.html#atan2(double,double)

将极角存入列表之后，再重复加入 + 2pi 的元素，这是为了计算完全。
最后使用滑动窗口计算。

class Solution {public int visiblePoints(List<List<Integer>> points, int angle, List<Integer> location) {int sameCnt = 0;List<Double> polarDegrees = new ArrayList<>();int x0 = location.get(0), y0 = location.get(1);for (int i = 0; i < points.size(); ++i) {int x = points.get(i).get(0), y = points.get(i).get(1);if (x == x0 && y == y0) {sameCnt++;continue;}Double degree = Math.atan2(y - y0, x - x0);polarDegrees.add(degree);}Collections.sort(polarDegrees);int m = polarDegrees.size();for (int i = 0; i < m; ++i) {polarDegrees.add(polarDegrees.get(i) + 2 * Math.PI);}int maxCnt = 0;double toDegree = angle * Math.PI / 180;for (int l = 0, r = 0; l < m; ++l) {Double curr = polarDegrees.get(l) + toDegree;while (r < polarDegrees.size() && polarDegrees.get(r) <= curr) {r++;}maxCnt = Math.max(maxCnt, r - l);}return maxCnt + sameCnt;}
}   

159. 至多包含两个不同字符的最长子串

https://leetcode.cn/problems/longest-substring-with-at-most-two-distinct-characters/description/

维护各个字符的出现次数，当哈希表中字符种类大于 2 的时候将左端点右移。

class Solution {public int lengthOfLongestSubstringTwoDistinct(String s) {int n = s.length(), ans = 0;Map<Character, Integer> cnt = new HashMap<>();for (int l = 0, r = 0; r < n; ++r) {cnt.merge(s.charAt(r), 1, Integer::sum);while (cnt.size() > 2) {cnt.merge(s.charAt(l), -1, Integer::sum);if (cnt.get(s.charAt(l)) == 0) cnt.remove(s.charAt(l));l++;}ans = Math.max(ans, r - l + 1);}return ans;}
}

340. 至多包含 K 个不同字符的最长子串

https://leetcode.cn/problems/longest-substring-with-at-most-k-distinct-characters/description/

把上一题的 2 换成 k 就好了。

class Solution {public int lengthOfLongestSubstringKDistinct(String s, int k) {int n = s.length(), ans = 0;Map<Character, Integer> cnt = new HashMap<>();for (int l = 0, r = 0; r < n; ++r) {cnt.merge(s.charAt(r), 1, Integer::sum);while (cnt.size() > k) {cnt.merge(s.charAt(l), -1, Integer::sum);if (cnt.get(s.charAt(l)) == 0) cnt.remove(s.charAt(l));l++;}ans = Math.max(ans, r - l + 1);}return ans;}
}