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.
Example:
Input: [0,1,0,3,12]
Output: [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.
题意:将数组中的0后置且不打乱非0数的顺序,要求不能创建数组
思路:用i寻找非0,赋值给前面对应非0的位置,index相当于是非0的位置,全部非0赋值完毕后再给数组的后面补齐对应个数的0
知识点:
1、++i和i++区别:
int i=0;
1)a =++i,先加再赋值,i =i+1=1,a=i=1;
2) a = i++;先赋值再加,a=i=0;i=i+1=1
2、看清是否需要return,函数类型是void则不需要return
class Solution {
public void moveZeroes(int[] nums) {
int len = nums.length;
int index = 0;
for(int i=0;i<len;i++) {
if(nums[i]!= 0){
nums[index++] = nums[i];
//相当于 num[index] = nums[i]; index +=1;
}
}
for(int i=index;i<len;i++){
nums[i] = 0;
}
// return nums; //函数void不需要return
}
}