LeetCode里马比赛的算法

大家好，欢迎来到IT知识分享网。

分为6组,定为a,b,c,d,e,f组，每组6匹马
第一轮，每组跑一次，取其冠军，假设每组的冠军是a1,b1,c1,d1,e1,f1
第二轮，将每组的冠军放一组里跑一次，取前3，假设为a1,b1,c1
(a1已为总冠军，无需再跑;c1是第3名，所以c1组其他成员无需再跑;d,e,f组所有成员也无需再跑)
第三轮，取a2, a3, b1, b2, b3, c1放一组里跑一次，取前2名,再加a1,即是最快的三匹马
总跑的次数:6+1+1=8

void HorseRace() { int horses[36]{}; for (int i = 0; i < 36; i++) { horses[i] = rand(); //随机马匹 } //第一轮:分为6组,进行6次比赛,horses[i*6]是每组的冠军 for (int i = 0; i < 6; i++) { std::sort(&horses[i * 6], &horses[i * 6 + 6], [](auto a, auto b) {return a > b; }); } int first_champion[6]; for (int i = 0; i < 6; i++) { first_champion[i] = horses[i * 6]; //取每组的冠军 } //第二轮:将每组的冠军比赛1次 std::sort(&first_champion[0], &first_champion[6], [](auto a, auto b) {return a > b; }); int final_champion[3]; //前3名的成绩 final_champion[0] = first_champion[0]; //总冠军，已经产生 int second_champion[6]; for (int i = 0; i < 6; i++) { if (first_champion[0] == horses[i * 6]) //找a1组 { //找到a2与a3 second_champion[0] = horses[i * 6 + 1]; second_champion[1] = horses[i * 6 + 2]; } else if (first_champion[1] == horses[i * 6]) //找b1组 { //找到b1 b2 b3 second_champion[2] = horses[i * 6 + 0]; second_champion[3] = horses[i * 6 + 1]; second_champion[4] = horses[i * 6 + 2]; } } second_champion[5] = first_champion[2]; //c1再次参与比赛 //第三轮:比赛1次 std::sort(&second_champion[0], &second_champion[6], [](auto a, auto b) {return a > b; }); //取前2名 final_champion[1] = second_champion[0]; final_champion[2] = second_champion[1]; //总成绩的前3名 for (int i = 0; i < 3; i++) { printf("%d ", final_champion[i]); } }