This is the first book of pleasure and delight 354 update, the first 379 Pian original
01 questions and look ready
Introduced today is the LeetCodearithmetic problem in Easythe first-level 216title (overall title number is 922). Given array A nonnegative integer, the integer is an odd number in half A, and the remaining half is even.
Sort the array, so that each time the A [i] is an odd number, I is an odd number; A [i] is an even number, I is an even number.
You can return any answers to meet an array of conditions. E.g:
Input: [4,2,5,7]
Output: [4,5,2,7]
Description: [4,7,2,5], [2,5,4,7], [2,7, 4,5] It will also be accepted.
Note :
2 <= A.length <= 20000
A.length％2 == 0
0 <= A [i] <= 1000
02 The first solution
Two Listodd, respectively even-save up to create a new array result, if the index is odd, odd on the storage Listmedium as the value of the new array element, contrary to the even-numbered memory Listin value as the new array element .
The time complexity of this solution is the O(N)space complexity Shi O(N).
public int[] sortArrayByParityII(int[] A) {
List<Integer> odd = new ArrayList<Integer>();
List<Integer> even = new ArrayList<Integer>();
for (int num : A) {
if (num%2 == 0) {
even.add(num);
} else {
odd.add(num);
}
}
int j = 0, k = 0;
int[] result = new int[A.length];
for (int i=0; i<result.length; i++) {
if (i%2 == 0) {
result[i] = even.get(j++);
} else {
result[i] = odd.get(k++);
}
}
return result;
}03 The second solution
We can also directly from Athe values, the same is to create a new resultarray to resultarray two new index, starting from a 0, only the even indices, the other starting from 1, only odd indices, twice traverse the array A, the index values and corresponding elements stored resultin.
The time complexity of this solution is the O(N)space complexity Shi O(N).
public int[] sortArrayByParityII2(int[] A) {
int[] result = new int[A.length];
int j = 0;
for (int i=0; i<A.length; i++) {
if (A[i]%2 == 0) {
result[j] = A[i];
j += 2;
}
}
int k = 1;
for (int i=0; i<A.length; i++) {
if (A[i]%2 != 0) {
result[k] = A[i];
k += 2;
}
}
return result;
}04 A third solution
The second solution for the above, we can also use only one cycle.
The time complexity of this solution is the O(N)space complexity Shi O(N).
public int[] sortArrayByParityII3(int[] A) {
int[] result = new int[A.length];
int j = 0, k = 1;
for (int i=0; i<A.length; i++) {
if (A[i]%2 == 0) {
result[j] = A[i];
j += 2;
} else {
result[k] = A[i];
k += 2;
}
}
return result;
}05 The fourth solution
Double pointer.
Define two pointers i and j, i representative of even index, starting with zero; J on behalf of the odd-indexed, beginning from n-1 (n is the length of the array A), if the index position corresponding to the even-odd elements, and the odd index positions corresponding element is an even number, the exchange proceeds element. If the index position corresponding to the even-numbered elements is an even number, the index i is incremented even number 2, similarly, the index position corresponding to the odd-odd elements, an odd index j is decremented by 2, the end of cycle condition j i is not less than 1 or less than n.
The time complexity of this solution is the O(N)space complexity Shi O(1).
public int[] sortArrayByParityII4(int[] A) {
int i = 0, j = A.length-1, n = A.length;
while (i < n && j >= 1) {
if (A[i]%2 == 1 && A[j]%2 == 0) {
int tem = A[j];
A[j] = A[i];
A[i] = tem;
}
if (A[i]%2 == 0) {
i += 2;
}
if (A[j]%2 == 1) {
j -= 2;
}
}
return A;
}06 Summary
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