記錄了初步解題思路 以及本地實(shí)現(xiàn)代碼；并不一定為最優(yōu) 也希望大家能一起探討 一起進(jìn)步

8/7 344. 反轉(zhuǎn)字符串

雙指針

def reverseString(s):
    """
    :type s: List[str]
    :rtype: None Do not return anything, modify s in-place instead.
    """
    l,r=0,len(s)-1
    while l<r:
        s[l],s[r]=s[r],s[l]
        l+=1
        r-=1

8/8 1749. 任意子數(shù)組和的絕對(duì)值的最大值

記錄最小值 最大值

def maxAbsoluteSum(nums):
    """
    :type nums: List[int]
    :rtype: int
    """
    n = len(nums)
    ans = 0
    minv,maxv = 0,0
    r = 0
    while r<n:
        minv+=nums[r]
        maxv+=nums[r]
        if maxv>0:
            ans = max(ans,maxv)
        else:
            maxv = 0
        if minv<0:
            ans = max(ans,-minv)
        else:
            minv = 0
        r+=1
    return ans

8/9 1281. 整數(shù)的各位積和之差

按要求計(jì)算

def subtractProductAndSum(n):
    """
    :type n: int
    :rtype: int
    """
    a,b = 1,0
    while n>0:
        v = n%10
        a *= v
        b += v
        n//=10
    return a-b

8/10 1289. 下降路徑最小和 II

從上到下遍歷每一行
在每一行中記錄最小值和次小值 及其位置
下一行使用到的必定只有這兩個(gè)值

def minFallingPathSum(grid):
    """
    :type grid: List[List[int]]
    :rtype: int
    """
    n = len(grid)
    first,second = (float("inf"),-1),(float("inf"),-1)
    for i in range(n):
        one,two = float("inf"),float("inf")
        f,s = -1,-1
        for j in range(n):
            cur = grid[i][j]
            if i!=0:
                if j==first[1]:
                    cur+=second[0]
                else:
                    cur+=first[0]
            if cur<one:
                two = one
                one = cur
                f = j
            elif cur<two:
                two = cur
                s=j
        first = (one,f)
        second = (two,s)
    return first[0]

8/11 1572. 矩陣對(duì)角線元素的和

計(jì)算每行需要累加的數(shù)位置

def diagonalSum(mat):
    """
    :type mat: List[List[int]]
    :rtype: int
    """
    ans = 0
    n= len(mat)
    for i in range(n):
        ans += mat[i][i]
        j = n-1-i
        if i!=j:
            ans+=mat[i][j]
    return ans

8/12 23. 合并 K 個(gè)升序鏈表

將每個(gè)鏈表當(dāng)前值放入小頂堆中
依次取堆中最小值 并將下一個(gè)節(jié)點(diǎn)放入

class ListNode(object):
     def __init__(self, val=0, next=None):
        self.val = val
        self.next = next
def mergeKLists(lists):
    """
    :type lists: List[ListNode]
    :rtype: ListNode
    """
    import heapq
    h = ListNode(0)
    cur = h
    l = []
    n = len(lists)
    for i in range(n):
        if lists[i]:
            heapq.heappush(l, (lists[i].val,i))
            lists[i]=lists[i].next
    while l:
        v,idx = heapq.heappop(l)
        cur.next = ListNode(v)
        cur = cur.next
        if lists[idx]:
            heapq.heappush(l,(lists[idx].val,idx))
            lists[idx]=lists[idx].next
    return h.next

8/13 88. 合并兩個(gè)有序數(shù)組

從大到小考慮文章來(lái)源地址http://www.zghlxwxcb.cn/news/detail-650003.html

def merge(nums1, m, nums2, n):
    """
    :type nums1: List[int]
    :type m: int
    :type nums2: List[int]
    :type n: int
    :rtype: None Do not return anything, modify nums1 in-place instead.
    """
    cur = m+n-1
    l,r=m-1,n-1
    while r>=0:
        while l>=0 and nums1[l]>nums2[r]:
            nums1[cur],nums1[l]=nums1[l],nums1[cur]
            cur-=1
            l-=1
        nums1[cur],nums2[r]=nums2[r],nums1[cur]
        cur-=1
        r-=1

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