python leetcode 410. Split Array Largest Sum

代码一：DP
难点在于如何设置dp数组
dp[i][j] 前j个数字分成i组所能得到的最小的各个子数组中最大值
代码二：二分查找变种

class Solution(object):
    def splitArray(self, nums, m):
        """
        :type nums: List[int]
        :type m: int
        :rtype: int
        """
        #dp[i][j] 前j个数字分成i组所能得到的最小的各个子数组中最大值
        n=len(nums)
        sums=[0]*(n+1)
        dp=[[float('inf')]*(n+1) for _ in range(m+1)]
        dp[0][0]=0
        for __ in range(1,n+1):
            sums[__]=sums[__-1]+nums[__-1]
            dp[1][__]=sums[__]
        for i in range(2,m+1):
            for j in range(1,n+1):
                for k in range(i-1,j):#不需要从0开始 因为已经分割了i-1次即如果分割后每个数组只有一个元素的话也有i-1个数组                 
                    val=max(dp[i-1][k],sums[j]-sums[k])#开辟出的新的子数组和是sum[j]-sum[k] 与之前的进行比较
                    dp[i][j]=min(dp[i][j],val)
        return dp[m][n]

class Solution:
    def splitArray(self, nums, m):
        """
        :type nums: List[int]
        :type m: int
        :rtype: int
        """
        left=max(nums)
        right=sum(nums)
        while left < right:
            mid=left+(right-left)//2
            if self.canSplit(nums,m,mid):
                right = mid 
            else:
                left=mid+1 
        return left
    def canSplit(self,nums,m,sum1):
        cnt=1
        curSum=0
        for i in range(len(nums)):
            curSum+=nums[i]
            if curSum>sum1:
                curSum=nums[i]
                cnt+=1
                if cnt>m: return False 
        return True