No11.week11 maxSubArray
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1.suscription
2.How to Slove
Hint:
consider contiguous subsequences ending exactly at position j.
Solve:
记f(j) = 以aj为结尾的最大和。
f(0) = 0
f(1) = a1
f(2) = max{ f(1) , f(1) + a2}
...
f(j) = max{ aj, f(j - 1)+ aj} // optimal substructure
算法复杂度:
时间复杂度:分别以每个数字为结尾,一共n个数字,故n个循环。每个循环做一个比较,f(j - 1), f(j - 1)+ aj两个谁比较大,复杂度为1,因此算法的时间复杂度为O(n)。
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