279. Perfect Squares

Given a positive integer n, find the least number of perfect square numbers (for example,1, 4, 9, 16, ...) which sum to n.

For example, given n = 12, return3 because 12 = 4 + 4 + 4; given n = 13, return2 because 13 = 4 + 9.

dp[0] = 0 dp[1] = dp[0]+1 = 1 dp[2] = dp[1]+1 = 2 dp[3] = dp[2]+1 = 3 dp[4] = Min{ dp[4-1*1]+1, dp[4-2*2]+1 }       = Min{ dp[3]+1, dp[0]+1 }       = 1                 dp[5] = Min{ dp[5-1*1]+1, dp[5-2*2]+1 }       = Min{ dp[4]+1, dp[1]+1 }       = 2 dp[13] = Min{ dp[13-1*1]+1, dp[13-2*2]+1, dp[13-3*3]+1 }        = Min{ dp[12]+1, dp[9]+1, dp[4]+1 }        = 2 dp[n] = Min{ dp[n - i*i] + 1 },  n - i*i >=0 && i >= 1

public int numSquares(int n) { int dp[] = new int[n + 1]; for (int i = 1; i <= n; i++) { dp[i] = i; // 最多都由1组成 for (int j = 1; j * j <= i; j++) // 0-1背包 dp[i] = Math.min(dp[i - j * j] + 1, dp[i]); } return dp[n]; }