写在前面:待补充。。。
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:
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1.You must do this in-place without making a copy of the array.
2.Minimize the total number of operations.
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我的解答:
class Solution {
public:
void moveZeroes(vector<int>& nums) {
int nums_length = nums.size();
int i = nums_length - 2;
int j = 0;
for(i; i >= 0; i--){
if (nums[i] == 0) {
for(j = i; j < (nums_length-1); j++){
nums[j] = nums[j+1];
}
nums[nums_length - 1] = 0;
}
}
}
};
不用vector,用普通数组,以下是完整代码
#include <iostream>
using namespace std;
int main(int argc, const char * argv[]) {
int nums[9] = {1,0,7,9,0,6,78,0,99};
int nums_length = sizeof(nums) / sizeof(nums[0]);
int i = nums_length - 2;
int j = 0;
int k = 0;
for(i; i >= 0; i--){
if (nums[i] == 0) {
for(j = i; j < (nums_length-1); j++){
nums[j] = nums[j+1];
}
nums[nums_length - 1] = 0;
}
}
for (k = 0; k < nums_length; k++) {
cout << nums[k] << " ";
}
return 0;
}