leetcode 1043 Partition Array for Maximum Sum 详解
这是个典型的DP问题,所以先分拆成小的子问题。
如:arr = [1,15, 7, 9, 2, 5, 10], k = 3
设f(i)f(i)f(i)存储的是当前数组arr[0,..,i−1]arr[0,..,i-1]arr[0,..,i−1]的解
f(0)=0f(0)=0f(0)=0 (1)
i=0,arr[0]=1,则f(1)=1i = 0, arr[0]=1, 则f(1)=1i=0,arr[0]=1,则f(1)=1 (2)
i=1,arr[1]=15,f(2)=max(f(0)+2∗max(arr[0,.,1]),f(1)+1∗arr[1])=30i=1, arr[1]=15, f(2) = max(f(0)+2*max(arr[0,.,1]), f(1)+1*arr[1])=30i=1,arr[1]=15,f(2)=max(f(0)+2∗max(arr[0,.,1]),f(1)+1∗arr[1])=30 (3)
i=2,arr[2]=7,则f(3)=max(f(0)+3∗max(arr[0,.,2]),f(1)+2∗max(arr[1,.,2]),f(2)+arr[2])=40i=2, arr[2]=7, 则f(3)=max(f(0)+3*max(arr[0,.,2]), f(1)+2*max(arr[1,.,2]), f(2)+arr[2])=40i=2,arr[2]=7,则f(3)=max(f(0)+3∗max(arr[0,.,2]),f(1)+2∗max(arr[1,.,2]),f(2)+arr[2])=40 (4)
i=3,arr[3]=9,则f(4)=max(f(1)+3∗max(arr[1,.,3]),f(2)+2∗max(arr[2,.,3]),f(3)+arr[3])=54i=3, arr[3] = 9, 则f(4) = max(f(1)+3*max(arr[1,.,3]), f(2)+2*max(arr[2,.,3]),f(3)+arr[3]) = 54i=3,arr[3]=9,则f(4)=max(f(1)+3∗max(arr[1,.,3]),f(2)+2∗max(arr[2,.,3]),f(3)+arr[3])=54 (5)
...
...
从上述例子中可以得出一个方程式:
f(i)=max(f(i−j)+j∗cur_max)f(i) = max(f(i-j) + j * cur\_max) f(i)=max(f(i−j)+j∗cur_max)j∈[1,min(i,k)]j \in [1, min(i,k)]j∈[1,min(i,k)]
最后代码如下: