283. 移动零

题目：https://leetcode-cn.com/problems/move-zeroes/

我的想法：使用一个慢指针和一个快指针，慢指针则是用来寻找数组中“0”的位置，快指针用来遍历整个数组以找到不为“0”的元素，然后交互

代码：

C++版本：
void moveZeroes(vector<int>& nums) {
    int fastIndex = 0, slowIndex = 0;
    while ( fastIndex < nums.size()){
        if (nums[slowIndex] == 0 && nums[fastIndex] != 0){
            nums[slowIndex] = nums[fastIndex];
            nums[fastIndex] = 0;
            slowIndex++;
        }
        else if (nums[slowIndex] != 0){
                slowIndex++;
        }
        fastIndex++;
    }
}

 

JAVA版本：    
public void moveZeroes(int[] nums) {
    int fastIndex = 0, slowIndex = 0;
    while ( fastIndex < nums.length){
            if (nums[slowIndex] == 0 && nums[fastIndex] != 0){
                nums[slowIndex] = nums[fastIndex];
                nums[fastIndex] = 0;
                slowIndex++;
            }
            else if (nums[slowIndex] != 0){
                slowIndex++;
            }
            fastIndex++;
     }
 }

 

标签：slowIndex,nums,++,fastIndex,int,283,移动,指针
来源： https://www.cnblogs.com/wltree/p/15392183.html