Commit ea59335

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update 5 leetcode

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	`1`	`+/*`
	`2`	`+ * @lc app=leetcode id=191 lang=javascript`
	`3`	`+ *`
	`4`	`+ * [191] Number of 1 Bits`
	`5`	`+ */`
	`6`	`+`
	`7`	`+// @lc code=start`
	`8`	`+/**`
	`9`	`+ * 按位与的思路,一直翻转翻转`
	`10`	`+ * @param {number} n - a positive integer`
	`11`	`+ * @return {number}`
	`12`	`+ */`
	`13`	`+var hammingWeight = function(n) {`
	`14`	`+ let res = 0;`
	`15`	`+ while (n) {`
	`16`	`+ n &= n - 1;`
	`17`	`+ res++;`
	`18`	`+ }`
	`19`	`+ return res;`
	`20`	`+};`
	`21`	`+// @lc code=end`

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	`1`	`+/*`
	`2`	`+ * @lc app=leetcode id=205 lang=javascript`
	`3`	`+ *`
	`4`	`+ * [205] Isomorphic Strings`
	`5`	`+ */`
	`6`	`+`
	`7`	`+// @lc code=start`
	`8`	`+/**`
	`9`	`+ * 维护哈希表,属性值对`
	`10`	`+ * 如果不匹配,就返回 false`
	`11`	`+ * @param {string} s`
	`12`	`+ * @param {string} t`
	`13`	`+ * @return {boolean}`
	`14`	`+ */`
	`15`	`+var isIsomorphic = function(s, t) {`
	`16`	`+ let mapS = {};`
	`17`	`+ let mapT = {};`
	`18`	`+ for (let i = 0; i < s.length; i++) {`
	`19`	`+ const si = s[i];`
	`20`	`+ const ti = t[i];`
	`21`	`+ // "badc" "baba" 因为有这种场景,所以需要保存两者的键值对来比较`
	`22`	`+ if ((mapS[si] && mapS[si] !== ti) \|\| (mapT[ti] && mapT[ti] !== si)) {`
	`23`	`+ return false;`
	`24`	`+ }`
	`25`	`+ mapS[si] = ti;`
	`26`	`+ mapT[ti] = si;`
	`27`	`+ }`
	`28`	`+ return true;`
	`29`	`+};`
	`30`	`+// @lc code=end`

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	`1`	`+/*`
	`2`	`+ * @lc app=leetcode id=231 lang=javascript`
	`3`	`+ *`
	`4`	`+ * [231] Power of Two`
	`5`	`+ */`
	`6`	`+`
	`7`	`+// @lc code=start`
	`8`	`+/**`
	`9`	`+ * @param {number} n`
	`10`	`+ * @return {boolean}`
	`11`	`+ */`
	`12`	`+var isPowerOfTwo = function(n) {`
	`13`	`+ if (n <= 1) {`
	`14`	`+ return n >= 0 ? !!n : false;`
	`15`	`+ }`
	`16`	`+`
	`17`	`+ while (n > 1) {`
	`18`	`+ if (n % 2 !== 0) {`
	`19`	`+ return false;`
	`20`	`+ }`
	`21`	`+ n /= 2;`
	`22`	`+ }`
	`23`	`+ return true;`
	`24`	`+};`
	`25`	`+`
	`26`	`+/**`
	`27`	`+ * 按位与的思路,二进制不同位都是 1 才会是 1`
	`28`	`+ * 而 2 的幂的二进制除了首位是 1 ,其他位都是 0`
	`29`	`+ * @param {*} n`
	`30`	`+ * @returns`
	`31`	`+ */`
	`32`	`+var isPowerOfTwoBetter = function(n) {`
	`33`	`+ return n > 0 && (n & (n - 1)) === 0;`
	`34`	`+};`
	`35`	`+`
	`36`	`+// @lc code=end`

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	`1`	`+/*`
	`2`	`+ * @lc app=leetcode id=258 lang=javascript`
	`3`	`+ *`
	`4`	`+ * [258] Add Digits`
	`5`	`+ */`
	`6`	`+`
	`7`	`+// @lc code=start`
	`8`	`+/**`
	`9`	`+ * @param {number} num`
	`10`	`+ * @return {number}`
	`11`	`+ */`
	`12`	`+var addDigits = function(num) {`
	`13`	`+ num = num.toString(); // 38`
	`14`	`+ let sum = 0;`
	`15`	`+ // 如果超过一位则继续往下走`
	`16`	`+ while (num.length > 1) {`
	`17`	`+ // 叠加位数`
	`18`	`+ for (let s of num) {`
	`19`	`+ sum += +s;`
	`20`	`+ console.info('sum', s, sum, num);`
	`21`	`+ }`
	`22`	`+ // 注意处理的时候返回字符串,重置总的和`
	`23`	`+ num = sum.toString();`
	`24`	`+ sum = 0;`
	`25`	`+ }`
	`26`	`+ return +num;`
	`27`	`+};`
	`28`	`+`
	`29`	`+var addDigitsBetter = function(num) {`
	`30`	`+ // 数学课上讲过如果各位数相加等于 9 就一定能被 9 整除`
	`31`	`+ // 这里各位数相加得到最后的一位数之内,也肯定是 9 的余数`
	`32`	`+ // 但 9 的情况特别,先借位再补位`
	`33`	`+ return (num - 1) % 9 + 1;`
	`34`	`+};`
	`35`	`+addDigits(38);`
	`36`	`+// @lc code=end`

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	`1`	`+/*`
	`2`	`+ * @lc app=leetcode id=342 lang=javascript`
	`3`	`+ *`
	`4`	`+ * [342] Power of Four`
	`5`	`+ */`
	`6`	`+`
	`7`	`+// @lc code=start`
	`8`	`+/**`
	`9`	`+ * 接着 2 的幂的按位与运算`
	`10`	`+ * 但是给出取反的二进制位`
	`11`	`+ * 1010101010...`
	`12`	`+ * 转成 16 进制提升可读性`
	`13`	`+ * @param {number} n`
	`14`	`+ * @return {boolean}`
	`15`	`+ */`
	`16`	`+var isPowerOfFour = function(n) {`
	`17`	`+ return n > 0 && (n & (n - 1)) === 0 && (n & 0xAAAAAAAA) === 0;`
	`18`	`+};`
	`19`	`+`
	`20`	`+/**`
	`21`	`+ * 4 的幂一定是 2 的幂`
	`22`	`+ * 二项式定理展开得知 4 的幂一定是除以 3 余数是 1`
	`23`	`+ * @param {*} n`
	`24`	`+ * @returns`
	`25`	`+ */`
	`26`	`+var isPowerOfFour2 = function(n) {`
	`27`	`+ return n > 0 && (n & (n - 1)) === 0 && (n % 3 === 1);`
	`28`	`+};`
	`29`	`+// @lc code=end`

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