LeetCode practice using python language
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#001. Two Sum
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#004. Median of Two Sorted Arrays
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#015. 3Sum
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#016. 3Sum Closest
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#018. 4Sum
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#026. Remove Duplicates from Sorted Array
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#027. Remove Element
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#0
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#0
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#035. Search Insert Position
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#0
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#039. Combination Sum
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#041. First Missing Positive
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#0
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#053. Maximum Subarray
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#055. Jump Game
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#0
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#066. Plus One
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#0
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#442. Find All Duplicates in an Array
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#485. Max Consecutive Ones
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#561. Array Partition I
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#566. Reshape the Matrix
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#667. Beautiful Arrangement II
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#766. Toeplitz Matrix
LeetCode link: https://leetcode.com/problems/array-partition-i/description/
Problem description:
Given an array of 2n integers, your task is to group these integers into n pairs of integer, say (a1, b1), (a2, b2), ..., (an, bn) which makes sum of min(ai, bi) for all i from 1 to n as large as possible.
Thought:
every other number starting from the second largest number is adopted.
LeetCode link: https://leetcode.com/problems/reshape-the-matrix/description/
Thought:
- sum(,[])
- zip(*([iter(flat)]*c))
- 运行到return直接结束
LeetCode link: https://leetcode.com/problems/find-all-duplicates-in-an-array/description/
Thought:
所有元素都在1~len(nums)之间，所以所有元素减一都可以对应到nums的某个index上。 遍历每个元素，其减一作为index标记对应元素负号，由于所有元素都是正的，对应到的元素如果是负的，说明之前标记过， 即本元素是重复的，则append到结果中
LeetCode link: https://leetcode.com/problems/max-consecutive-ones/description/
Thought:
My answer is to change the original list and don't forget to refer the former index. Other solution renews the maximum number every iteration.
LeetCode link: https://leetcode.com/problems/beautiful-arrangement-ii/description/
Thought:
Thought: When k = n-1, a valid construction is [1, n, 2, n-1, 3, n-2, ....]. One way to see this is, we need to have a difference of n-1, which means we need 1 and n adjacent; then, we need a difference of n-2, etc.
This leads to the following idea: we will put [1, 2, ...., n-k-1] first, and then we have N = k+1 adjacent numbers left, of which we want k different differences. This is just the answer above translated by n-k-1: we'll put [n-k, n, n-k+1, n-1, ....] after
LeetCode link: https://leetcode.com/problems/remove-element/description/
Thought:
我的做法是利用了语言特性。但不如第二个参考答案用的好。首先remove(int)按照顺序删去指定int，count(obj)可以得出list中obj的个数。 参考答案一更机智。通过设定起始index，在遇到val时把当下数字和end对换，start点不变，end点前移，leave a 'val' behind.不遇到val则start点前移， 直到end超过start时取nums中前start个数字
LeetCode link: https://leetcode.com/problems/combination-sum/description/
Thought:
LeetCode link: https://leetcode.com/problems/toeplitz-matrix/description/
Thought:
Think really carefully about the relationship between i and j.
LeetCode link: https://leetcode.com/problems/two-sum/description/
Thought:
LeetCode link: https://leetcode.com/problems/median-of-two-sorted-arrays/description/
**Thought: **
hard. Use binary search. Note the median is used for dividing a set into two equal length subsets, that one subset is always greater than the other
LeetCode link: https://leetcode.com/problems/3sum/description/
Thought: carefully deal with duplicate; use 'left' and 'right' index to narrow the inner loop.
LeetCode link: https://leetcode.com/problems/3sum-closest/description/
Thought: same idea as 015. 3Sum, left and right index is a useful way to reduce time from N^3 to N^2
LeetCode link: https://leetcode.com/problems/4sum/description/
Thought: pass
LeetCode link: Thought: pass
LeetCode link: https://leetcode.com/problems/jump-game/description/
Thought: 当碰到0的时候判断现在的maximum jump是否大于尾数，maximum jump想象成汽油，你能驶过的距离是现在已经驶过的距离加上这一站的汽油量nums[i]，记录行程中这个汽油量的最大值。
LeetCode link: https://leetcode.com/problems/search-insert-position/description/
Thought: easy
LeetCode link: https://leetcode.com/problems/first-missing-positive/description/
Thought: 好好使用set的功能
LeetCode link: https://leetcode.com/problems/plus-one/description/
Thought: easy
LeetCode link: https://leetcode.com/problems/maximum-subarray/description/
Thought: O(n)tims-Kadane's algorithm https://en.wikipedia.org/wiki/Maximum_subarray_problem A bit of a background: Kadane's algorithm is based on splitting up the set of possible solutions into mutually exclusive (disjoint) sets. It exploits the fact that any solution (i.e., any member of the set of solutions) will always have a last element {\displaystyle i} i (this is what is meant by "sum ending at position {\displaystyle i} i"). Thus, we simply have to examine, one by one, the set of solutions whose last element's index is {\displaystyle 1} 1, the set of solutions whose last element's index is {\displaystyle 2} 2, then {\displaystyle 3} 3, and so forth to {\displaystyle n} n. It turns out that this process can be carried out in linear time. Kadane's algorithm begins with a simple inductive question: if we know the maximum subarray sum ending at position {\displaystyle i} i (call this {\displaystyle B_{i}} B_{i}), what is the maximum subarray sum ending at position {\displaystyle i+1} i+1 (equivalently, what is {\displaystyle B_{i+1}} B_{{i+1}})? The answer turns out to be relatively straightforward: either the maximum subarray sum ending at position {\displaystyle i+1} i+1 includes the maximum subarray sum ending at position {\displaystyle i} i as a prefix, or it doesn't (equivalently, {\displaystyle B_{i+1}=max(A_{i+1},A_{i+1}+B_{i})} {\displaystyle B_{i+1}=max(A_{i+1},A_{i+1}+B_{i})}, where {\displaystyle A_{i+1}} A_{i+1} is the element at index {\displaystyle i+1} i+1).
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