阿华代码，不是逆风，就是我疯
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目录
一：长度最小的子数组
二：无重复字符的最长子串
三：最大连续1的个数
四：将x减到0的最小操作数
五：水果成篮
六：找到字符串中所有字母的异位词
七：串联所有单词的子串
八：最小覆盖子串
一：长度最小的子数组
class Solution {public int minSubArrayLen(int target, int[] nums) {int left = 0 , right = 0 , sum = 0 , n = nums.length;int len = Integer.MAX_VALUE;for(;right < n ; right++){//进入窗口sum += nums[right];while(sum >= target){//更新值len = Math.min(len,right - left + 1);//出窗口sum -= nums[left];left++;}}return len == Integer.MAX_VALUE ? 0 : len; }
}二：无重复字符的最长子串
3. 无重复字符的最长子串 - 力扣（LeetCode）
class Solution {public int lengthOfLongestSubstring(String ss) {int left = 0 , right = 0 , ret = 0;char[] s = ss.toCharArray();int n = s.length;int[] hash = new int[128];while(right < n){hash[s[right]]++;while(hash[s[right]] > 1){hash[s[left++]]--;//出窗口}ret = Math.max(ret , right - left + 1);right++;}return ret;}
}三：最大连续1的个数
class Solution {public int longestOnes(int[] nums, int k) {int left = 0 , right = 0 , count = 0;int n = nums.length , ret = 0;while(right < n){if(nums[right] == 1){right++;}else{count++;right++;}while(count > k){if(nums[left] == 1){left++;}else{count--;left++;}}ret = Math.max(ret,right-left);} return ret;}
}四：将x减到0的最小操作数
1658. 将 x 减到 0 的最小操作数 - 力扣（LeetCode）
class Solution4 {public int minOperations(int[] nums, int x) {int left = 0 , right = 0 , count = 0 ,sum = 0;int n = nums.length;for(int i = 0 ; i < n ; i++){sum += nums[i];}int target = sum - x , tem = 0;if(target < 0){return -1;}if(target == 0){return n;}for(; right < n ; right++){//进窗口不判断tem += nums[right];while(tem > target ){tem -= nums[left];left++;}//执行顺序也有讲究，最后一步判断if(tem == target){count = Math.max(count , right-left+1);}}if(count == 0){return -1;}return n-count;}
}五：水果成篮
904. 水果成篮 - 力扣（LeetCode）
class Solution {public int totalFruit(int[] f) {Map<Integer,Integer> hash = new HashMap<Integer,Integer>();int ret = 0;for(int left = 0 , right = 0 ; right < f.length ; right++){//进窗口int in = f[right];hash.put(in,hash.getOrDefault(in,0)+1);while(hash.size() > 2){//判断//出窗口int out = f[left];hash.put(out,hash.get(out)-1);if(hash.get(out) == 0){hash.remove(out);}left++;}ret = Math.max(ret,right - left + 1);}return ret;}
}六：找到字符串中所有字母的异位词
438. 找到字符串中所有字母异位词 - 力扣（LeetCode）
class Solution {public List<Integer> findAnagrams(String ss, String pp) {List<Integer> list = new ArrayList<Integer>();char[] s = ss.toCharArray();char[] p = pp.toCharArray();int[] hashS = new int[26];int[] hashP = new int[26];for(char x : p){hashP[ x - 'a']++;} for(int right = 0 , left = 0 , count = 0 ; right < ss.length() ; right++ ){char in = s[right];hashS[ in - 'a']++;if(hashS[in - 'a'] <= hashP[in - 'a']){count++;//进入的是有效字符}char out = s[left];if(right - left + 1 > pp.length()){hashS[out - 'a']--;if(hashS[out - 'a'] < hashP[out - 'a']){count--;}left++;}if(count == pp.length()){list.add(left);}}return list;}
}七：串联所有单词的子串
30. 串联所有单词的子串 - 力扣（LeetCode）
class Solution8 {public List<Integer> findSubstring(String s, String[] words) {List<Integer> ret = new ArrayList<Integer>();//先把words中的元素放到HashMap中//每个元素的长度为m，数组长度为n,记录元素种类eleMap<String,Integer> hash1 = new HashMap<String,Integer>();int m = words[0].length() , n = words.length ,ele = 0;for(String ss : words){hash1.put(ss,hash1.getOrDefault(ss,0)+1);}for(int i = 0 ; i < m ; i++ ){//wc太绝了hashMap每次i移动时需要初始化一下，要不然上一次的值还存留在HashMap中Map<String,Integer> hash2 = new HashMap<String,Integer>();for(int left = i , right = i , count = 0 ; right + m <= s.length() ; right += m){//进窗口String in = s.substring(right,right+m);hash2.put(in,hash2.getOrDefault(in,0)+1);//判断count加不加if(hash2.get(in) <= hash1.getOrDefault(in,0)){count++;}//出窗口while(right - left + 1 > m * n ){String out = s.substring(left,left+m);left+=m;if(hash2.get(out) <= hash1.getOrDefault(out,0)){count--;}hash2.put(out,hash2.get(out)-1);}if(count == n){ret.add(left);}}}return ret;}}
八：最小覆盖子串
76. 最小覆盖子串 - 力扣（LeetCode）
class Solution {public String minWindow(String ss, String tt) {//先把要找的字符tt转化为数组并放到哈希表里char[] t = tt.toCharArray();int[] hash1 = new int[128];int kind = 0;//统计tt字符串中有多少种字符for(char ch : t){hash1[ch]++;if(hash1[ch] == 1){kind++;}}//同样把字符ss转化为数组char[] s = ss.toCharArray();int[] hash2 = new int[128];int len = Integer.MAX_VALUE;int left = 0 , right = 0 , begin = -1 ;for(int count = 0; right < s.length ; right++){//进窗口char in = s[right];hash2[in]++;//判断如果直接判断两个哈希表非常耗费时间引入countif(hash1[in] == hash2[in]){count++;}//更新结果(如果种类一直相同，那就一直出窗口所以用while)while(count == kind){if(right - left + 1 < len){begin = left;len = right-left+1;}//出窗口char out = s[left++];if(hash2[out] == hash1[out] ){//考虑两种情况，出的是有效字符还是无效字符count--;}hash2[out]--; }}//for循环走完了一直进不去while循环，返回空字符串if(len == Integer.MAX_VALUE){return new String();}else{String ret = ss.substring(begin,begin+len);return ret;}}
}