移动零

283. 移动零 - 力扣（LeetCode）

看到这道题，是否第一个想法就是双指针，下面我们就来使用双指针来完成这道题.

图:

代码:

java:

 // 移动 0 双指针
    public void moveZeroes(int[] nums) {
        for (int current = 0, dest = -1; current < nums.length; current++) {
            if (nums[current] != 0) {
                // 此时不等于 0 进行交换
                dest++;
                swapNumber(nums, current, dest);
            }
        }
    }

    public void swapNumber(int[] numbers, int left, int right) {
        int temp = numbers[left];
        numbers[left] = numbers[right];
        numbers[right] = temp;
    }

python:

class Solution(object):
    def moveZeroes(self, nums):
        current = 0
        dest = -1
        while len(nums):
            if nums[current] != 0:
                dest = dest + 1
                self.sawpNumbs(nums, current, dest)
            current = current + 1
            if current >= len(nums):
                break

    def sawpNumbs(self, nums, left, right):
        temp = nums[left]
        nums[left] = nums[right]
        nums[right] = temp

这里双指针，划分数组如果学过快排，应该会有映像.

复写零

仔细阅读题目，应该会有一个想法就是双指针, 一个进行遍历，一个进行复写,下面来看看思路.

图一:

图二:

java

  public static void duplicateZeros(int[] arr) {
            int current = 0, dest = -1;
            while (current < arr.length) {
                if (arr[current] == 0) {
                    dest += 2;
                } else {
                    dest++;
                }
                if (dest >= arr.length - 1) {
                    // 此时说明 current 到达了最后一个复写元素
                    break;
                }
                current++;
            }
            if (dest == arr.length) {
                // 处理一下边界情况
                arr[arr.length - 1] = 0;
                current--;
                dest = dest - 2;
            }

            // 进行赋值操作
            while (current >= 0) {
                if (arr[current] != 0) {
                    arr[dest--] = arr[current];
                } else {
                    arr[dest--] = 0;
                    arr[dest--] = 0;
                }
                current--;
            }
    }

python

class Solution(object):
    def duplicateZeros(self, arr):
        current = 0
        dest = -1
        while current < len(arr):
            if arr[current] == 0:
                dest += 2
            else:
                dest += 1
            if dest >= len(arr) - 1:
                break
            current += 1
        # 判断边界
        if dest == len(arr):
            arr[len(arr) - 1] = 0
            dest = dest - 2
            current = current - 1
        while current >= 0:
            if arr[current] != 0:
                arr[dest] = arr[current]
                dest = dest - 1
            else:
                arr[dest] = 0
                arr[dest - 1] = 0
                dest = dest - 2
            current = current - 1