【Leetcode Medium】1318. Minimum Flips to Make a OR b Equal to c

原题链接

题目描述

Given 3 positives numbers a, b and c. Return the minimum flips required in some bits of a and b to make ( a OR b == c ). (bitwise OR operation).
Flip operation consists of change any single bit 1 to 0 or change the bit 0 to 1 in their binary representation.

修改a或者b的二进制表示，使得a or b = c

解题思路

尝试做了一场leetcode的周赛题目，这是第二题（直接倒在了第三题），题目不难，我的思路也非常常规，就是把三个数的二进制值都get到然后进行比较，一个小坑是三个数的二进制表示长度可能不一，要提前处理（由于题目约束里说了三个数都是正数，所以在前面加0就可以）
code：

class Solution {public int minFlips(int a, int b, int c) {String C = Integer.toBinaryString(c);String A = Integer.toBinaryString(a);String B = Integer.toBinaryString(b);int flip = 0;String[] addtmp = addZero(A, B, C);char[] binaryA = addtmp[0].toCharArray();char[] binaryB = addtmp[1].toCharArray();char[] binaryC = addtmp[2].toCharArray();   for (int i = binaryC.length - 1; i >= 0; i--) {char tmp = binaryC[i];char tmpA = binaryA[i];char tmpB = binaryB[i];//if tmp == 1, requires one or two '1' in A and B//if tmp == 0, requires both 0 in A and Bif (tmp == '1') {if (tmpA == '0' && tmpB == '0') {flip++;}}else {if ((tmpA == '1' && tmpB == '0') || (tmpA == '0' && tmpB == '1')) {flip++;}else if (tmpA == '1' && tmpB == '1')flip += 2;}}return flip;}private String[] addZero(String a, String b, String c) {int maxLen = a.length();if (b.length() > maxLen)maxLen = b.length();if (c.length()> maxLen)maxLen = c.length();int tmp = maxLen - a.length();String zerotmp = "";for (int i = 0; i < tmp; i++) {zerotmp += "0";}a = zerotmp + a;tmp = maxLen - b.length();zerotmp = "";for (int i = 0; i < tmp; i++) {zerotmp += "0";}b = zerotmp + b;tmp = maxLen - c.length();zerotmp = "";for (int i = 0; i < tmp; i++) {zerotmp += "0";}c = zerotmp + c;String[] res = new String[3];res[0] = a;res[1] = b;res[2] = c;return res;}
}

TIP：

对于我的解法来说，由于使用了一个private方法，由此引出了一个重要的知识点：
Java中函数传参是call by value
Java中所有的参数传递，传递的都是副本
如果原变量代表的是value（如基本类型），那么在函数中修改该值，对原变量不会造成影响；如果原变量代表的是地址（如String，或自定义对象类型），给副本重新赋值（即修改指向）不会改变原变量，若对副本指向的内容做出改变，则会影响到原变量
具体可以参考这一篇，写的比较详细
https://www.cnblogs.com/woshimrf/p/5263018.html

其他解法

本质思路是一样的，只不过利用二进制数的操作（如左移，做与操作判断是1还是0）可以避免长度不一致问题，就不需要我的那个private方法，也不需要转换成二进制再转成charArray这样了

class Solution {public int minFlips(int a, int b, int c) {int res = 0;for(int i = 1, j = 0; j < 32; i = i<< 1, j++) {//分别判断a，b，c的当前位是1 or 0int v1 = a & i;int v2 = b & i;int v3 = c & i;if(v3 != 0) {//c的当前位为1if(v1 == 0 && v2 == 0) {res++;}} else {if(v1 != 0 && v2 != 0) {res += 2;}else if(v1 != 0 || v2 != 0) {res++;}}    }return res;}
}