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2 changes: 1 addition & 1 deletion cpp/dp/maximum-subarray.md
Original file line number Diff line number Diff line change
Expand Up @@ -35,7 +35,7 @@ $$target = \max\left\{f[j]\right\}, 1 \leq j \leq n$$
* 思路2:直接在i到j之间暴力枚举,复杂度是`O(n^3)`
* 思路3:处理后枚举,连续子序列的和等于两个前缀和之差,复杂度`O(n^2)`。
* 思路4:分治法,把序列分为两段,分别求最大连续子序列和,然后归并,复杂度`O(nlog n)`
* 思路5:把思路2`O(n^2)`的代码稍作处理,得到`O(n)`的算法
* 思路5:把思路3`O(n^2)`的代码稍作处理,得到`O(n)`的算法
* 思路6:当成M=1的最大M子段和


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