Problem:
Solution:
class Solution {
public:
void moveZeroes(vector<int>& nums) {
int size = nums.size();
int iter = 0;
for (int zero_iter = 0; zero_iter < size; zero_iter++) {
if (nums[zero_iter] != 0) {
nums[iter++] = nums[zero_iter];
if (iter - 1 == zero_iter) {
// Iter == zero_iter means base case: no zero till now.
continue;
}
nums[zero_iter] = 0;
}
}
return;
}
};
===== Another =====
void moveZeroes(vector<int>& nums) {
int count = 0, iter = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] == 0) {
count++;
} else {
nums[iter++] = nums[i];
}
}
for (int i = iter; i < nums.size(); i++) {
nums[i] = 0;
}
return;
}
Given an array
nums
, write a function to move all 0
's to the end of it while maintaining the relative order of the non-zero elements.
For example, given
nums = [0, 1, 0, 3, 12]
, after calling your function, nums
should be [1, 3, 12, 0, 0]
.
Note:
- You must do this in-place without making a copy of the array.
- Minimize the total number of operations.
Solution:
class Solution {
public:
void moveZeroes(vector<int>& nums) {
int size = nums.size();
int iter = 0;
for (int zero_iter = 0; zero_iter < size; zero_iter++) {
if (nums[zero_iter] != 0) {
nums[iter++] = nums[zero_iter];
if (iter - 1 == zero_iter) {
// Iter == zero_iter means base case: no zero till now.
continue;
}
nums[zero_iter] = 0;
}
}
return;
}
};
===== Another =====
void moveZeroes(vector<int>& nums) {
int count = 0, iter = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] == 0) {
count++;
} else {
nums[iter++] = nums[i];
}
}
for (int i = iter; i < nums.size(); i++) {
nums[i] = 0;
}
return;
}
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