git是一种分布式代码管理工具，git通过树的形式记录文件的更改历史，比如： base'--base--A--A' ^ | --- B--B' 小米工程师常常需要寻找两个分支最近的分割点，即b...

// ConsoleApplication10.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include <iostream>
#include <vector>
using namespace std;class Solution {
public:/*** 返回git树上两点的最近分割点** @param matrix 接邻矩阵，表示git树，matrix[i][j] == '1' 当且仅当git树中第i个和第j个节点有连接，节点0为git树的跟节点* @param indexA 节点A的index* @param indexB 节点B的index* @return 整型*/int getSplitNode(vector<string> matrix, int indexA, int indexB) {if (indexA >= matrix.size() || indexB >= matrix.size()){return 0;}vector<int> depth;vector<int> parent;breadTraverl(matrix, depth, parent);int p=0;//如果同一深度，父节点相同，则分割点为父节点//同一深度，父节点不同，寻找父节点的父节点//不同深度，从深度高的往上过滤，每次过滤的时候判断是否达到低的节点的深度；没有之前，要判断低的节点是否是深的节点的父节点if (depth[indexA] == depth[indexB]){int a = indexA;int b = indexB;while (parent[a]!=parent[b]){a = parent[a];b = parent[b];}p= parent[a];}else {int a = indexA;//a对应的节点深int b = indexB;bool flag = false;//未找到分割点if (depth[indexA] < depth[indexB]){a = indexB;b = indexA;}while (depth[a]!=depth[b])//当节点a和节点b不在同一个深度{if (parent[a] == b){p = b;flag = true;break;}else{--a;}}if (flag == false){int a = indexA;int b = indexB;while (parent[a] != parent[b]){a = parent[a];b = parent[b];}p = parent[a];}}return p;}//广度优先遍历图void breadTraverl(vector<string> matrix, vector<int> &depth, vector<int> &parent){//visited数组，如何visited[i]为false，则被访问过vector<bool> visited;//  vector<int> depth;//  vector<int> parent;for (int i = 0;i < matrix.size();++i){visited.push_back(true);depth.push_back(0);parent.push_back(0);}cout << "V_" << 0 << "  ";visited[0] = false;depth[0] = 0;parent[0] = 0;//根节点的本身设置为自己for (int i = 0;i < matrix.size();++i){//  cout << "matrix[0].length():" << matrix[0].length() << endl;for (int j = 0;j < matrix[0].length();++j){//注意：此处的matrix[i][j]应该为'1'而不是1if ((matrix[i][j] == '1')&& (visited[j] == true)){cout << "matrix[" << i << "]"<<"["<<j<<"]:" << matrix[i][j] << "  ";cout << "visited["<<j<<"]:" << visited[j] << endl;cout << "V_" << j << "  ";parent[j] = i;depth[j] = depth[i] + 1;visited[j] = false;}}}cout << endl;for (int i = 0;i < matrix.size();i++){cout << i << ":";cout << "parent:" << parent[i] << "  ";cout << "depth:" << depth[i] << endl;}}};int main()
{Solution so;vector<string> matrix;string str1 = "01011";string str2 = "10100";string str3 = "01000";string str4 = "10000";string str5 = "10000";matrix.push_back(str1);matrix.push_back(str2);matrix.push_back(str3);matrix.push_back(str4);matrix.push_back(str5);cout<<"结果："<<so.getSplitNode(matrix, 0, 2);cout << endl;return 0;
}