Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n.less
Example 1:code
Input: n = 12 Output: 3 Explanation: 12 = 4 + 4 + 4.
Example 2:it
Input: n = 13 Output: 2 Explanation: 13 = 4 + 9.
難度: mediumio
題目:給定一正整數n,找出其最少個平方數之和爲n。ast
思路:參考 (322. Coin Change)class
Runtime: 22 ms, faster than 91.72% of Java online submissions for Perfect Squares.
Memory Usage: 36.6 MB, less than 47.46% of Java online submissions for Perfect Squares.im
class Solution { public int numSquares(int n) { int s = (int) Math.floor(Math.sqrt(n)); if (s * s == n) { return 1; } int[] squares = new int[s]; int[] nums = new int[n + 1]; for (int i = 1; i <= s; i++) { squares[i - 1] = i * i; if (squares[i - 1] < nums.length) { nums[squares[i - 1]] = 1; } } for (int i = squares[0]; i <= n; i++) { if (nums[i] > 0) { continue; } nums[i] = n + 1; for (int j = 0; j < squares.length && i - squares[j] >= 0; j++) { if (nums[i - squares[j]] > 0) { nums[i] = Math.min(nums[i - squares[j]] + 1, nums[i]); } } } return nums[n]; } }