content
1. LeetCode 228. Summary interval
topic
Given an ordered integer array nums with no repeating elements.
Returns a list of the smallest ordered interval ranges that exactly cover all the numbers in the array. That is, every element of nums is exactly covered by some interval range, and there is no number x that belongs to some range but not to nums.
Input: nums = [0,1,2,4,5,7]
Output: ["0->2","4->5","7"]
Xiaobian cooking solution
public static List<String> summaryRanges(int[] nums) {
int step = -1;
int next = 0;
int over = 0;
int curren = 0;
List<String> list = new ArrayList<>();
for(int i = 0;i<nums.length;i++){
int c = nums[i];
if (i<=curren + step){
continue;
}
curren = i;
String temp = "";
step = 0;
if(i+1+step == nums.length){
list.add(String.valueOf(c));
break;
}
next = nums[i+1+step];
while (next - c == 1+step){
over = next;
step++;
if(i+1+step < nums.length){
next = nums[i+1+step];
}
}
if (step == 0){
temp = String.valueOf(c);
}else{
temp += c + "->"+over;
}
list.add(temp);
}
return list;
}
Enough trouble, dishes.
Big boss pointing the country
public static List<String> summaryRanges2(int[] nums) {
List<String> ret = new ArrayList<String>();
int i = 0;
int n = nums.length;
while (i < n) {
int low = i;
i++;
while (i < n && nums[i] == nums[i - 1] + 1) {
i++;
}
int high = i - 1;
StringBuffer temp = new StringBuffer(Integer.toString(nums[low]));
if (low < high) {
temp.append("->");
temp.append(Integer.toString(nums[high]));
}
ret.add(temp.toString());
}
return ret;
}
2. Power of LeetCode 231.2
topic
Given an integer n, please judge whether the integer is a power of 2. If so, return true; otherwise, return false.
If there is an integer x such that n == 2x, n is considered to be a power of 2.
Xiaobian cooking solution
public static boolean isPowerOfTwo(int n) {
if (n == 1){
return true;
}
while (true){
if (n/2 >1 && n%2==0){
n = n/2;
}else{
if (n == 2){
return true;
}else{
return false;
}
}
}
}
Thought analysis
A number nn is a power of 22 if and only if nn is a positive integer and the binary representation of nn contains only 11 11s.
Therefore, we can consider using bit operation to extract the lowest 11 in the binary representation of nn, and then judge whether the remaining value is 00. n & (n - 1)
where \texttt{&}& means bitwise AND operation. This bit operation trick can directly remove the lowest bit 11 of the binary representation of nn.
Big boss pointing the country
public boolean isPowerOfTwo(int n) {
return n > 0 && (n & (n - 1)) == 0;
}
Binary is not very good at playing, ashamed ah.
3. LeetCode 205. Valid letter anagrams
topic
Given two strings s and t, write a function to determine whether t is an anagram of s.
Note: If each character in s and t appears the same number of times, then s and t are called anagrams of each other.
Xiaobian problem solving ideas
Take out the number of all letters and put them into the map. If the two strings are anagrams, the maps must be equal.
Xiaobian cooking solution
public static boolean isAnagram(String s, String t) {
Map<Character,Integer> map = new HashMap<>();
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if(map.containsKey(c)){
map.put(c,map.get(c)+1);
}else{
map.put(c,1);
}
}
Map<Character,Integer> map2 = new HashMap<>();
for (int i = 0; i < t.length(); i++) {
char c = t.charAt(i);
if(map2.containsKey(c)){
map2.put(c,map2.get(c)+1);
}else{
map2.put(c,1);
}
}
return map.equals(map2);
}
Thought analysis
t is an anagram of ss equivalent to "two strings sorted equal". Therefore, we can sort the strings ss and tt respectively, and judge whether the sorted strings are equal. Furthermore, if ss and tt are of different lengths, tt must not be an anagram of ss.
Big boss pointing the country
public static boolean isAnagram(String s, String t) {
char[] c1 = s.toCharArray();
char[] c2 = t.toCharArray();
Arrays.sort(c1);
Arrays.sort(c2);
return Arrays.equals(c1,c2);
}
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