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Add C++ solutions for leetcode 869.

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Lines changed: 34 additions & 0 deletions

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	`1`	`+#include<vector>`
	`2`	`+#include<algorithm>`
	`3`	`+#include<unordered_map>`
	`4`	`+#include<iostream>`
	`5`	`+using namespace std;`
	`6`	`+`
	`7`	`+class Solution {`
	`8`	`+public:`
	`9`	`+ bool reorderedPowerOf2(int n) {`
	`10`	`+ for (int i = 1; i <= 1e9; i *= 2)`
	`11`	`+ {`
	`12`	`+ if (canOrderToSame(i, n)) return true;`
	`13`	`+ }`
	`14`	`+ return false;`
	`15`	`+ }`
	`16`	`+ bool canOrderToSame(int i, int n) /* 判断整数i是否重新排序后能和n相等 */`
	`17`	`+ {`
	`18`	`+ unordered_map<char, int> dict1, dict2;`
	`19`	`+ for (auto ch : to_string(i)) dict1[ch]++; /* 转为字符串, 统计其中每一个字符出现的次数 */`
	`20`	`+ for (auto ch : to_string(n)) dict2[ch]++;`
	`21`	`+ return dict1 == dict2;`
	`22`	`+ }`
	`23`	`+};`
	`24`	`+`
	`25`	`+// Test`
	`26`	`+int main()`
	`27`	`+{`
	`28`	`+ Solution sol;`
	`29`	`+ int n = 24;`
	`30`	`+ auto res = sol.reorderedPowerOf2(n);`
	`31`	`+ cout << (res ? "True" : "False") << endl;`
	`32`	`+`
	`33`	`+ return 0;`
	`34`	`+}`

Lines changed: 48 additions & 0 deletions

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	`1`	`+#include<vector>`
	`2`	`+#include<algorithm>`
	`3`	`+#include<unordered_map>`
	`4`	`+#include<iostream>`
	`5`	`+using namespace std;`
	`6`	`+`
	`7`	`+class Solution {`
	`8`	`+public:`
	`9`	`+ bool reorderedPowerOf2(int n) {`
	`10`	`+ for (int i = 1; i <= 1e9; i *= 2)`
	`11`	`+ {`
	`12`	`+ if (canOrderToSame(i, n)) return true;`
	`13`	`+ }`
	`14`	`+ return false;`
	`15`	`+ }`
	`16`	`+ bool canOrderToSame(int i, int& n) /* 判断整数i是否重新排序后能和n相等 */`
	`17`	`+ {`
	`18`	`+ for (int d = 0; d <= 9; d++) /* 枚举每一位的数字(0~9之间), 只要出现某一个数字在整数i和n中的数量不相等就返回false */`
	`19`	`+ {`
	`20`	`+ int count1 = countTimes(i, d);`
	`21`	`+ int count2 = countTimes(n, d);`
	`22`	`+ if (count1 != count2) return false;`
	`23`	`+ }`
	`24`	`+ return true;`
	`25`	`+ }`
	`26`	`+ int countTimes(long int x, int d) /* 统计每一位的数digit(简写为d)在整数x中出现的次数 */`
	`27`	`+ {`
	`28`	`+ int count = 0;`
	`29`	`+ while (x)`
	`30`	`+ {`
	`31`	`+ if (x % 10 == d)`
	`32`	`+ count++;`
	`33`	`+ x = x / 10;`
	`34`	`+ }`
	`35`	`+ return count;`
	`36`	`+ }`
	`37`	`+};`
	`38`	`+`
	`39`	`+// Test`
	`40`	`+int main()`
	`41`	`+{`
	`42`	`+ Solution sol;`
	`43`	`+ int n = 24;`
	`44`	`+ auto res = sol.reorderedPowerOf2(n);`
	`45`	`+ cout << (res ? "True" : "False") << endl;`
	`46`	`+`
	`47`	`+ return 0;`
	`48`	`+}`

Lines changed: 48 additions & 0 deletions

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	`1`	`+#include<vector>`
	`2`	`+#include<algorithm>`
	`3`	`+#include<unordered_map>`
	`4`	`+#include<iostream>`
	`5`	`+using namespace std;`
	`6`	`+`
	`7`	`+class Solution {`
	`8`	`+public:`
	`9`	`+ bool reorderedPowerOf2(int n) {`
	`10`	`+ for (int i = 1; i <= 1e9; i *= 2)`
	`11`	`+ {`
	`12`	`+ if (canOrderToSame(i, n)) return true;`
	`13`	`+ }`
	`14`	`+ return false;`
	`15`	`+ }`
	`16`	`+ bool canOrderToSame(int i, int& n) /* 判断整数i是否重新排序后能和n相等 */`
	`17`	`+ {`
	`18`	`+ vector<int> dict1(11), dict2(11);`
	`19`	`+ for (int d = 0; d <= 9; d++)`
	`20`	`+ {`
	`21`	`+ dict1[d] = countTimes(i, d);`
	`22`	`+ dict2[d] = countTimes(n, d);`
	`23`	`+ }`
	`24`	`+ return dict1 == dict2;`
	`25`	`+ }`
	`26`	`+ int countTimes(long int x, int d) /* 计算每一位的数digit(简写为d)在整数x中出现的次数 */`
	`27`	`+ {`
	`28`	`+ int count = 0;`
	`29`	`+ while (x)`
	`30`	`+ {`
	`31`	`+ if (x % 10 == d)`
	`32`	`+ count++;`
	`33`	`+ x = x / 10;`
	`34`	`+ }`
	`35`	`+ return count;`
	`36`	`+ }`
	`37`	`+};`
	`38`	`+`
	`39`	`+// Test`
	`40`	`+int main()`
	`41`	`+{`
	`42`	`+ Solution sol;`
	`43`	`+ int n = 24;`
	`44`	`+ auto res = sol.reorderedPowerOf2(n);`
	`45`	`+ cout << (res ? "True" : "False") << endl;`
	`46`	`+`
	`47`	`+ return 0;`
	`48`	`+}`

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