Move Zeros
題意：
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.
解題思路：
updated on 2016.1.17
public void moveZeroes(int[] nums) {
int j = 0;
for(int i = 0; i < nums.length; i++) {
if(nums[i] != 0) {
int temp = nums[j];
nums[j] = nums[i];
nums[i] = temp;
j++;
}
}
}
public void moveZeroes(int[] nums) {
int j = 0; // The index of the leftmost zero in nums.
for(int i = 0; i < nums.length; i++){
if(nums[i] != 0){
if(i > j){ // i can only be larger than or equal to j.
nums[j] = nums[i];
nums[i] = 0;
}
j++;
}
}
}
使用兩根指針，一個不斷指向陣列中最靠前的0元素，另一根從該指針的後面去找一個元零元素，最後再將兩根指針元素交換即可，其程式碼如下：
public class Solution {
public void moveZeroes(int[] nums) {
if (nums == null || nums.length == 0) {
return;
}
for (int i = 0, j = 0; i < nums.length; i++) {
if (nums[i] == 0) {
j = i + 1;
while (j < nums.length && nums[j] == 0) {
j++;
}
if (j < nums.length) {
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
}
}
}
}