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Merge pull request #1712 from c0dedance/patch-1

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Update 0343.整数拆分.md
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程序员Carl
2022-11-04 17:22:20 +08:00
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parent ce64a3f754 5c3a57ea1c
commit fe16fac72f
1 changed files with 6 additions and 6 deletions
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problems/0343.整数拆分.md
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@ -89,7 +89,7 @@ dp[i] 是依靠 dp[i - j]的状态，所以遍历i一定是从前向后遍历，
所以遍历顺序为：
```
for (int i = 3; i <= n ; i++) {
for (int j = 1; j < i - 1; j++) {
for (int j = 1; j <= i / 2; j++) {
dp[i] = max(dp[i], max((i - j) * j, dp[i - j] * j));
}
}
@ -110,7 +110,7 @@ public:
vector<int> dp(n + 1);
dp[2] = 1;
for (int i = 3; i <= n ; i++) {
for (int j = 1; j < i - 1; j++) {
for (int j = 1; j <= i / 2; j++) {
dp[i] = max(dp[i], max((i - j) * j, dp[i - j] * j));
}
}
@ -167,7 +167,7 @@ public:
dp[2] = 2;
dp[3] = 3;
for (int i = 4; i <= n ; i++) {
for (int j = 1; j < i - 1; j++) {
for (int j = 1; j <= i / 2; j++) {
dp[i] = max(dp[i], dp[i - j] * dp[j]);
}
}
@ -224,7 +224,7 @@ class Solution:
# 假设对正整数 i 拆分出的第一个正整数是 j（1 <= j < i），则有以下两种方案：
# 1) 将 i 拆分成 j 和 i−j 的和，且 i−j 不再拆分成多个正整数，此时的乘积是 j * (i-j)
# 2) 将 i 拆分成 j 和 i−j 的和，且 i−j 继续拆分成多个正整数，此时的乘积是 j * dp[i-j]
for j in range(1, i - 1):
for j in range(1, i / 2 + 1):
dp[i] = max(dp[i], max(j * (i - j), j * dp[i - j]))
return dp[n]
```
@ -281,7 +281,7 @@ var integerBreak = function(n) {
dp[2] = 1
for(let i = 3; i <= n; i++) {
for(let j = 1; j < i; j++) {
for(let j = 1; j <= i / 2; j++) {
dp[i] = Math.max(dp[i], dp[i - j] * j, (i - j) * j)
}
}
@ -306,7 +306,7 @@ function integerBreak(n: number): number {
const dp: number[] = new Array(n + 1).fill(0);
dp[2] = 1;
for (let i = 3; i <= n; i++) {
for (let j = 1; j <= i - 2; j++) {
for (let j = 1; j <= i / 2; j++) {
dp[i] = Math.max(dp[i], j * dp[i - j], j * (i - j));
}
}