From ed9ebe1c4e894575100ecdafe9bfaf5ad698814a Mon Sep 17 00:00:00 2001 From: Luo <82520819+Jerry-306@users.noreply.github.com> Date: Tue, 28 Sep 2021 09:05:09 +0800 Subject: [PATCH] =?UTF-8?q?=E4=BF=AE=E6=94=B9=200077=20=E7=BB=84=E5=90=88?= =?UTF-8?q?=E4=BC=98=E5=8C=96=20c++=E4=BB=A3=E7=A0=81=20=E6=B2=A1=E6=9C=89?= =?UTF-8?q?=E6=A0=B7=E5=BC=8F=E9=97=AE=E9=A2=98?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0077.组合优化.md | 10 +++++----- 1 file changed, 5 insertions(+), 5 deletions(-) diff --git a/problems/0077.组合优化.md b/problems/0077.组合优化.md index 136ceb34..ff447bc1 100644 --- a/problems/0077.组合优化.md +++ b/problems/0077.组合优化.md @@ -22,7 +22,7 @@ 大家先回忆一下[77. 组合]给出的回溯法的代码: -``` +```c++ class Solution { private: vector> result; // 存放符合条件结果的集合 @@ -54,7 +54,7 @@ public: 在遍历的过程中有如下代码: -``` +```c++ for (int i = startIndex; i <= n; i++) { path.push_back(i); backtracking(n, k, i + 1); @@ -78,7 +78,7 @@ for (int i = startIndex; i <= n; i++) { **如果for循环选择的起始位置之后的元素个数 已经不足 我们需要的元素个数了,那么就没有必要搜索了**。 注意代码中i,就是for循环里选择的起始位置。 -``` +```c++ for (int i = startIndex; i <= n; i++) { ``` @@ -100,13 +100,13 @@ for (int i = startIndex; i <= n; i++) { 所以优化之后的for循环是: -``` +```c++ for (int i = startIndex; i <= n - (k - path.size()) + 1; i++) // i为本次搜索的起始位置 ``` 优化后整体代码如下: -``` +```c++ class Solution { private: vector> result;