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
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
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
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
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ele
Map<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
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count
if(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;
}
}
}