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]
.
给定一个数组的数组,写一个函数将所有0的结束它在保持非零元素的相对顺序。
例如,给定次数= [ 0,1,0,3,12 ],呼唤你的功能后,你应该[ 1,3,12,0,0 ]。
方法1:
class Solution {
public:
void moveZeroes(vector<int>& nums) {
int length = nums.size();
int i = 0;
while (i < length){
if(nums[i] == 0){
for(int j = i;j<length-1;j++){
nums[j] = nums[j+1];
}
nums[length-1] = 0;
length--;
}else{
i++;
}
}
}
};
方法2:
class Solution {
public:
void moveZeroes(vector<int>& nums) {
int index = 0;
//把非0元素压缩到数组前边
for(int i=0;i<nums.size();i++){
if(nums[i] != 0){
nums[index] = nums[i];
index++;
}
}
//剩下的全部置为0
for(;index<nums.size();index++){
nums[index] = 0;
}
}
};