Problem
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 1 :
大体思路是利用两个指针 p0 和 p1 从前往后扫, p0去找为零的, p1去找非零的,找到之后交换下。
需要注意已经排序好的test case , 如{ 1,0 }
class Solution {
public:
void moveZeroes(vector<int>& nums) {
const int N = nums.size();
for(int p0 = 0, p1 = 0; p0 < N && p1 < N;){
if(nums[p0] != 0) {p0++; <span style="color:#ff0000;">p1 = p0;</span> continue;} // p1 需要在 p0后面,否则{1,0}不通过
if(nums[p1] == 0) {p1++; continue;} // find none zero
swap(nums[p0++], nums[p1++]);
}
}
};Solution 2 :
code更加简洁,思路一致
class Solution {
public:
void moveZeroes(vector<int>& nums) {
const int N = nums.size();
for( int p0 = 0, p1 = 0; p1 < N; p1++){
if(nums[p1]) {
swap(nums[p0++], nums[p1]);
}
}
}
};
本文介绍了一种高效的算法,用于将数组中的所有零元素移动到数组的末尾,同时保持非零元素的相对顺序不变。该算法采用双指针技术实现原地操作,避免了额外的数据复制,减少了操作次数。
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