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June brush question 01 - array
2022-07-06 09:38:00 【A Guang chasing dreams】
Brush questions in June 01—— Array
Today's brush topic content : Array
Preface
- Update the problem solution content of the problem brush every day
- Focus on personal understanding , Look at the difficulty and update the number of questions
- The title comes from Li Kou
- Try to work out at least one question every day
- Language java、python、c\c++
One 、 Today's topic
- 1588. The sum of all odd length subarrays
- 1848. Minimum distance to target element
- 1652. Dismantle the bomb
- 1640. Can you join to form an array
Two 、 Their thinking
1. 1588. The sum of all odd length subarrays
- Require the sum of all odd subsequences
- That is, start traversing from each element , Add the current element sum to the result when it is an odd number of elements
- Return the result
class Solution {
public int sumOddLengthSubarrays(int[] arr) {
int i = 0, j, count = 0, ans = 0;
int n = arr.length;
while( i < n){
j = i;
count = 0;
while(j < n){
count += arr[j];
if ((j - i + 1) % 2 == 1){
ans += count;
}
j++;
}
i++;
}
return ans;
}
}
2. 1848. Minimum distance to target element
- Find the subscript of the element , Maintain a minimum value
class Solution {
public int getMinDistance(int[] nums, int target, int start) {
int i;
int n = nums.length;
int ret = 10010;
for (i = 0; i < n; ++i){
if (nums[i] == target){
ret = Math.min(ret, Math.abs(i - start));
}
}
return ret;
}
}
3. 1652. Dismantle the bomb
- according to
k
There are three cases for the value of , For the sake of understanding , Directly mark the three situations- If
k
by 0 Back to full 0 that will do- If
k
Greater than or less than 0, There are two scenarios , The expansion here is larger than 0 Of
- If
i+k
Greater than the rightmost subscript of the array , It will return to the first... From the beginning of the arrayi + k - n
A place- If it is less than the rightmost subscript of the array , Then add it directly
- Less than 0 It's the same thing
class Solution {
public int[] decrypt(int[] code, int k) {
int i, n = code.length;
int t = k;
int[] ret = new int[n];
if (k == 0){
return ret;
}else if (k < 0){
for (i = 0; i < n; ++i){
int temp = 0;
while(k < 0){
if (i + k < 0){
temp += code[n + i + k];
}else{
temp += code[i + k];
}
k++;
}
k = t;
ret[i] = temp;
}
}else{
for (i = 0; i < n; ++i){
int temp = 0;
while(k > 0){
if (i + k > n - 1){
temp += code[i + k - n];
}else{
temp += code[i + k];
}
k--;
}
k = t;
ret[i] = temp;
}
}
return ret;
}
}
4. 1640. Can you join to form an array
There are two ways of thinking about this question
Train of thought
The first is to use hash table
1, Convert all subsets of the two-dimensional array into strings as
key
In the hash table , Defaultvalue
byfalse
2. Traversing a one-dimensional array , If the current number exists in the... Of the hash tablekey
in , Willvalue
Set astrue
3. Splice the current numbers in order , Judge whether there is inkey
in , If it exists, repeat step 2
class Solution {
public boolean canFormArray(int[] arr, int[][] pieces) {
// Define a hash table
HashMap<String, Boolean> map = new HashMap<>();
int i = 0, j, n = arr.length;
// Convert all subsets in the two-dimensional array into strings and store them in the hash table
for(int[] item: pieces){
String s = Integer.toString(item[0]);
for(int k = 1; k < item.length; k++){
s += Integer.toString(item[k]);
}
map.put(s, false); // By default value Set as false
}
while(i < n){
String s = Integer.toString(arr[i]);
// If it is a single subset ( There's only one number )
if (map.containsKey(s)){
map.put(s, true);
}else{
// Is a subset of multiple numbers
for(j = i+1; j < n; j++){
s += Integer.toString(arr[j]);
if (map.containsKey(s)){
map.put(s, true);
i = j;
break;
}
}
}
i++;
}
// Traverse to see if there is still false Value
for (Boolean b: map.values()){
if (!b) return false;
}
return true;
}
}
Train of thought two :
- Traversing a one-dimensional array , Use one flag Identify the return result
- Find an equal number from the second array , It must be arranged in order
- according to flag Judge the result
class Solution {
public boolean canFormArray(int[] arr, int[][] pieces) {
int i = 0, j, k;
boolean flag;
while (i < arr.length){
flag = false;
for(j = 0; j < pieces.length; j++){
// If the first is not equal to arr[i] Then skip
if (pieces[j][0] != arr[i]){
continue;
}
// First equals arr[i] when , Judge pieces[j] Whether the subset of and arr The next number corresponds to
for(k = 0; k < pieces[j].length; k++){
if (pieces[j][k] == arr[i]){
i++;
flag = true;
}else{
return false;
}
}
// If at this time i The traversal is complete
if (i == arr.length){
return true;
}
}
if(!flag){
return false;
}
}
return true;
}
}
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