From a2a2484f00c0a2a806c8a3810c261e04cbf47058 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Jilang=20Miao=20=E7=BC=AA=E4=BD=B6=E6=9C=97?= Date: Sun, 25 Jun 2017 22:54:37 -0400 Subject: [PATCH] typo --- cpp/dp/maximum-subarray.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/cpp/dp/maximum-subarray.md b/cpp/dp/maximum-subarray.md index 47b9c53db..2fd5f742e 100644 --- a/cpp/dp/maximum-subarray.md +++ b/cpp/dp/maximum-subarray.md @@ -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