【Leetcode】Integer Break
2016-05-27 23:46
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题目链接:https://leetcode.com/problems/integer-break/
题目:
Given a positive integer n, break it into the sum of at least two positive integers and maximize the product of those integers. Return the maximum product you can get.
For example, given n = 2, return 1 (2 = 1 + 1); given n = 10, return 36 (10 = 3 + 3 + 4).
Note: you may assume that n is not less than 2.
思路:
找规律,n>=7时,c[i]=c[i-3]*3
算法:
public int integerBreak(int n) {
int res6[] = new int[] { 0, 0, 1, 2, 4, 6, 9 };
if (n <= 6)
return res6
;
int res[] = new int[n + 1];
for (int i = 2; i <= 6; i++) {
res[i] = res6[i];
}
for (int i = 7; i <= n; i++) {
res[i] = res[i - 3] * 3;
}
return res
;
}
题目:
Given a positive integer n, break it into the sum of at least two positive integers and maximize the product of those integers. Return the maximum product you can get.
For example, given n = 2, return 1 (2 = 1 + 1); given n = 10, return 36 (10 = 3 + 3 + 4).
Note: you may assume that n is not less than 2.
思路:
找规律,n>=7时,c[i]=c[i-3]*3
算法:
public int integerBreak(int n) {
int res6[] = new int[] { 0, 0, 1, 2, 4, 6, 9 };
if (n <= 6)
return res6
;
int res[] = new int[n + 1];
for (int i = 2; i <= 6; i++) {
res[i] = res6[i];
}
for (int i = 7; i <= n; i++) {
res[i] = res[i - 3] * 3;
}
return res
;
}
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