LeetCode[15] - Flip Game II
2015-11-03 本文已影响729人
土汪
这个题目李特是屌炸天的。
我飞了九牛二虎之力(路子对),但是代码写的七荤八素,好长好长好长好长的。
结果正解,三四行就搞定了。真是心有不甘啊。
想法如下:
保证p1能胜利,就必须保持所有p2的move都不能赢。
同时,p1只要在可走的Move里面,有一个move可以赢就足够了。(题目里面用一个for loop + 只要 满足条件就return true来表达 OR的意思:p1不同的路子,�赢一�种就行了)
p1: player1
p2: player2
/*
You are playing the following Flip Game with your friend: Given a string that contains only these two characters: + and -, you and your friend take turns to flip two consecutive "++" into "--". The game ends when a person can no longer make a move and therefore the other person will be the winner.
Write a function to determine if the starting player can guarantee a win.
For example, given s = "++++", return true. The starting player can guarantee a win by flipping the middle "++" to become "+--+".
Follow up:
Derive your algorithm's runtime complexity.
Tags: Backtracking
Similar Problems: (E) Nim Game, (E) Flip Game
*/
/*
Attemp2, from:http://www.cnblogs.com/jcliBlogger/p/4886741.html
Similar to my idea, but much more clear: no need of the isP1 flag.
Iterative idea:p1 can win, and p2 must not win at all.
Therefore, if p2's move can't win, we return true on p1's state.
For loop and the if statement works as 'OR': just need one of the p1's movement win.
*/
public class Solution {
public static boolean canWin(String s) {
if (s == null || s.length() <= 1) {
return false;
}
String str = new String(s);
for (int i = str.indexOf("++"); i >= 0 && i < str.length() - 1; i = str.indexOf("++")) {
if (!canWin( s.substring(0, i) + "--" + s.substring(i + 2))) {//Just pick one way of p1's move
return true;
}
str = str.substring(0, i) + "-" + str.substring(i + 1);//Help to move certain spot.
}
return false;
}
}
//let k = n/2
//O(k * (k - 1) * (k - 2) ... k) = O(k!) = O((n/2)!) = O(n!)
/*
Attempt1, Failed.
Thoughts:
method checkP1Win(), inside of it:
OR all p1's win state, if one of the move wins, return true;
However, a bit of code redundancy, does not feel good about this.
Fails on "+++++++++"
*/
public class Solution {
public static boolean canWin(String s) {
if (s == null || s.length() <= 1) {
return false;
}
boolean rst = false;
String str = new String(s);
for (int i = str.indexOf("++"); i >= 0 && i < str.length() - 1; i = str.indexOf("++")) {
if (checkP1Win(s, i, true)) {
rst = true;
break;
}
str = str.substring(0, i) + "-" + str.substring(i + 1);
}
return rst;
}
public static boolean checkP1Win(String str, int x, boolean isP1) {
String s = str.substring(0,x) + "--" + str.substring(x + 2);
if (s.indexOf("++") == -1) {
return isP1;
}
for (int i = s.indexOf("++"); i >= 0 && i < s.length() - 1; i = s.indexOf("++")) {
if (checkP1Win(s, i, !isP1)) {
return true;
}
s = s.substring(0, i) + "-" + s.substring(i + 1);
}
return false;
}
}
public class Solution {
public static boolean canWin(String s) {
if (s == null || s.length() <= 1) {
return false;
}
boolean rst = false;
String str = new String(s);
for (int i = str.indexOf("++"); i >= 0 && i < str.length() - 1; i = str.indexOf("++")) {
if (checkP1Win(s, i, true)) {
rst = true;
break;
}
str = str.substring(0, i) + "-" + str.substring(i + 1);
}
return rst;
}
public static boolean checkP1Win(String str, int x, boolean isP1) {
String s = str.substring(0,x) + "--" + str.substring(x + 2);
if (s.indexOf("++") == -1) {
return isP1;
}
if (isP1) {
String temp = s;
for (int i = temp.indexOf("++"); i >= 0 && i < temp.length() - 1; i = temp.indexOf("++")) {
if (checkP1Win(s, i, !isP1)) {
return true;
}
temp = temp.substring(0, i) + "-" + temp.substring(i + 1);
}
return false;
} else {
String temp = s;
for (int i = temp.indexOf("++"); i >= 0 && i < temp.length() - 1; i = temp.indexOf("++")) {
if (!checkP1Win(s, i, !isP1)) {
return false;
}
temp = temp.substring(0, i) + "-" + temp.substring(i + 1);
}
return true;
}
}
}