Python list in-place exchange nums[i], nums[nums[i]] = nums[nums[i]], nums[i] solution

Problem Description

Python list in-place exchange problem, when the index is in 数组[索引]the expression form , inappropriate exchange operation will lead to wrong output

1. Method 1 (wrong)

nums[i], nums[nums[i]] = nums[nums[i]], nums[i]

2. Method 2 (Correct)

nums[nums[i]], nums[i] = nums[i], nums[nums[i]]

test

# 法1
nums1 = [3,3,2,2,1,1]
i = 0
nums1[i], nums1[nums1[i]] = nums1[nums1[i]], nums1[i]
print(nums1)

# 法2
nums2 = [3,3,2,2,1,1]
i = 0
nums2[nums2[i]], nums2[i] = nums2[i], nums2[nums2[i]]
print(nums2)

执行完成，耗时：44 ms
[2, 3, 3, 2, 1, 1]
[2, 3, 2, 3, 1, 1]

reason

During execution , the left and right will be formed into tuplesnums1[i], nums1[nums1[i]] = nums1[nums1[i]], nums1[i] in python , and then assigned according to the principle of first left and then right . Therefore, it is advisable to set ( a , b ) = ( c , d ) (a, b) = (c, d)$(a,b)=(c,d)$ , the process is$a=c,b=d$ , so here$nums1[i]=When\; nums1\left[nums1\right[i\left]\right]$ , nums [ i ] nums[i]has been changed$The\; value\; of\; nums\left[i\right]$ , then$nums1[nums1[i]]=When\; nums1\left[i\right]$ , the index on the left side of the equation changes, so an error occurs.

Specifically, $i=0$ , first execute$nums1[0]=nums1\left[nums1\right[0\left]\right]$ , so
$$nums1[nums1[0]]=nums1[3]=2$$$$nums1\left[0\right]=2$$
Execute again$nums1[nums1[0]]=When\; nums1\left[0\right]$ , the right side of the equation is calculated first, and then the value is copied to the left side, as follows:
$$nums1[0]=2$$$$nums1\left[nums1\right[0\left]\right]=nums1[2]=2$$

Therefore, the value modified in the second step nums1[2]is not modified, and nums1[3]the result is[2, 3, 3, 2, 1, 1]

think

For the correct way, no detailed analysis here