[LeetCode]283. Move Zeroes&26. Remove Duplicates from Sorted Array

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本文介绍如何在不使用额外空间的情况下高效地处理数组中的元素，包括将零移至数组末尾及去除重复项的方法。

283 . Move Zeroes
Easy

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.

0ms:

public void moveZeroes(int[] nums) {
    int lastNonZeroFoundAt = 0;
    for(int i=0;i<nums.length;i++){
        if (nums[i]!=0) 
            nums[lastNonZeroFoundAt++]=nums[i];
    }     
    for(int i=lastNonZeroFoundAt;i<nums.length;i++){
        nums[i] = 0;
    }
}

1ms:

public void moveZeroes(int[] nums) {
         int[] cord = new int[nums.length];
         int p = 0;
         for(int i=0;i<nums.length;i++){
            cord[i]=p;
            if(nums[i]==0)
                p++;
         }
         for(int j=0;j<cord.length;j++){
            int phase = cord[j];
            nums[j-phase]=nums[j];
         }
         for(int k=p;k>0;k--){
             nums[nums.length-k]=0;
         }
}

26 . Remove Duplicates from Sorted Array

Easy

Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.

Do not allocate extra space for another array, you must do this in place with constant memory.

For example,
Given input array nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. It doesn’t matter what you leave beyond the new length.

12ms:

public int removeDuplicates(int[] nums) {
         int p = 0;
         for(int i=1;i<nums.length;i++){
             if(nums[i]!=nums[p])
                 nums[++p] = nums[i];
         }
         return 1+p;
}