LeetCode-283 Move Zeroes
2016-06-14 23:38
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https://leetcode.com/problems/move-zeroes/
Given an array
to the end of it while maintaining the relative order of the non-zero elements.
For example, given
be
Note:
You must do this in-place without making a copy of the array.
Minimize the total number of operations.
1、遍历数组,并统计0的个数count,对于下标为i的非0的元素,移至下标为i-count的位置;最终数组末尾补上count个0。
(16ms)
class Solution {
public:
void moveZeroes(vector<int>& nums) {
int len = nums.size(),count=0,i=0;
while(i<len){
while(nums[i]==0 && i<len){
count++;
i++;
}
while(nums[i]!=0 && i<len){
nums[i-count]=nums[i];
i++;
}
}
for(i = len - count;i<len;i++){
nums[i] = 0;
}
}
};
2、遍历数组,删除0元素,即计算每个非0元素是数组中的第几个非0元素就将其放在该位置,最终补上0。
(20ms)
class Solution {
public:
void moveZeroes(vector<int>& nums) {
int len = nums.size(),count=0;
for(int i = 0;i<len;i++){
if(nums[i]!=0)
nums[count++]=nums[i];
}
for(int i = count;i<len;i++)
nums[i] = 0;
}
};
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,
numsshould
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.
1、遍历数组,并统计0的个数count,对于下标为i的非0的元素,移至下标为i-count的位置;最终数组末尾补上count个0。
(16ms)
class Solution {
public:
void moveZeroes(vector<int>& nums) {
int len = nums.size(),count=0,i=0;
while(i<len){
while(nums[i]==0 && i<len){
count++;
i++;
}
while(nums[i]!=0 && i<len){
nums[i-count]=nums[i];
i++;
}
}
for(i = len - count;i<len;i++){
nums[i] = 0;
}
}
};
2、遍历数组,删除0元素,即计算每个非0元素是数组中的第几个非0元素就将其放在该位置,最终补上0。
(20ms)
class Solution {
public:
void moveZeroes(vector<int>& nums) {
int len = nums.size(),count=0;
for(int i = 0;i<len;i++){
if(nums[i]!=0)
nums[count++]=nums[i];
}
for(int i = count;i<len;i++)
nums[i] = 0;
}
};
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