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Create Find the Punishment Number of an Integer

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	`1`	`+想法:對每個 1 <= i <= n 都測試是否是 punushment number,使用一個遞迴找過所有可能組合;如果任一組合符合條件就輸出 true,否則輸出 false`
	`2`	`+對每個數字都檢查並把最後答案相加即可`
	`3`	`+`
	`4`	`+Time Complexity : O(n * 2^(log_10n)) for every number needs O(2^(log_10 n) ) time to check and there are n numbers need to check`
	`5`	`+Space Complexity : O(2^(log_10n)) for the recursion (recursion depth = O(logn))`
	`6`	`+`
	`7`	`+class Solution {`
	`8`	`+public:`
	`9`	`+ bool valid(int num , int target) {`
	`10`	`+ if (target == num )`
	`11`	`+ return true ;`
	`12`	`+ if ( target <= 0 \|\| num == 0 )`
	`13`	`+ return false ;`
	`14`	`+ for(int power = 10 ; power < num ; power *= 10) {`
	`15`	`+ if ( valid(num / power , target - (num % power)) )`
	`16`	`+ return true ;`
	`17`	`+ }`
	`18`	`+ return false ;`
	`19`	`+ }`
	`20`	`+ int punishmentNumber(int n) {`
	`21`	`+ int ans = 0 ;`
	`22`	`+ for(int i = 1 ; i <= n ; i++) {`
	`23`	`+ if ( valid(i * i , i) )`
	`24`	`+ ans += i * i ;`
	`25`	`+`
	`26`	`+ }`
	`27`	`+ return ans ;`
	`28`	`+ }`
	`29`	`+};`

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