[LeetCode] Move Zeroes
阿新 • • 發佈:2018-12-21
Given an array
nums
, write a function to move all0
'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
題意:把陣列的零移到末尾,其餘元素相對位置不變,不能使用新陣列,讓操作次數儘可能小。
思路:用到一個佇列,從前向後列舉,每當遇到一個零,向佇列裡新增這個零元素的位置,每當遇到非零元素,看是否佇列不為空,不為空的話就取出一個元素,把這個非零元素放在從佇列裡取出來的對應位置上,然後把這個非零元素的位置加入佇列,並把這個位置的元素改為零。
C程式碼:
void moveZeroes(int* nums, int numsSize) { int head,tail; head = tail = 0; int i,pos[10005]; for(i = 0; i < numsSize; i++) { if(nums[i] == 0) { pos[tail++] = i; } else if(head <4 tail) { pos[tail++] = i; nums[pos[head]] = nums[i]; head++; nums[i] = 0; } } }
Java程式碼:
public class Solution { public void moveZeroes(int[] nums) { int head,tail,i,len; int[] queue = new int[20005]; len = nums.length; head = tail = 0; for(i = 0; i < len; i++) { if(nums[i] == 0) { queue[tail++] = i; } else if(head < tail){ queue[tail++] = i; nums[queue[head]] = nums[i]; head++; nums[i] = 0; } } } }