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.
Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.
把一个数组里所有的0都移到后面，不能改变非零数的相对位置关系，而且不能拷贝额外的数组。
要用替换法in-place，双指针来做，前面的指针指向处理过的最后一个非零数字， 后面一个指针向后扫，遇到非零数字，和前面指针所指数字交换位置，前面指针后移1位。
或者不交换只把非零的数字换到前面，最后子统一把后面的全部变成0。
Java: Time Complexity: O(n), Space Complexity: O(1)
public class Solution {
public void moveZeroes(int[] nums) {
int index = 0;
for (int i = 0; i < nums.length; ++i) {
if (nums[i] != 0) {
nums[index++] = nums[i];
}
}
for (int i = index; i < nums.length; ++i) {
nums[i] = 0;
}
}
}
Python:
class Solution(object):
def moveZeroes(self, nums):
"""
:type nums: List[int]
:rtype: void Do not return anything, modify nums in-place instead.
"""
pos = 0
for i in xrange(len(nums)):
if nums[i]:
nums[i], nums[pos] = nums[pos], nums[i]
pos += 1
C++:
class Solution {
public:
void moveZeroes(vector<int>& nums) {
int pos = 0;
for (auto& num : nums) {
if (num) {
swap(nums[pos++], num);
}
}
}
};
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