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Merge branch 'master' of github.com:youngyangyang04/leetcode-master

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programmercarl
2022-11-08 09:45:50 +08:00
parent 730dd3c76f 4df81ae0bb
commit 0ad33e5743
2 changed files with 17 additions and 17 deletions
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10
problems/0343.整数拆分.md
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@ -88,7 +88,7 @@ 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： 所以遍历顺序为：
```CPP ```CPP
for (int i = 3; i <= n ; i++) { 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)); dp[i] = max(dp[i], max((i - j) * j, dp[i - j] * j));
} }
} }
@ -191,7 +191,7 @@ public:
dp[2] = 2; dp[2] = 2;
dp[3] = 3; dp[3] = 3;
for (int i = 4; i <= n ; i++) { 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]); dp[i] = max(dp[i], dp[i - j] * dp[j]);
} }
} }
@ -248,7 +248,7 @@ 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： # 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) # 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] # 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])) dp[i] = max(dp[i], max(j * (i - j), j * dp[i - j]))
return dp[n] return dp[n]
``` ```
@ -305,7 +305,7 @@ var integerBreak = function(n) {
dp[2] = 1 dp[2] = 1
for(let i = 3; i <= n; i++) { 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) dp[i] = Math.max(dp[i], dp[i - j] * j, (i - j) * j)
} }
} }
@ -330,7 +330,7 @@ function integerBreak(n: number): number {
const dp: number[] = new Array(n + 1).fill(0); const dp: number[] = new Array(n + 1).fill(0);
dp[2] = 1; dp[2] = 1;
for (let i = 3; i <= n; i++) { 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)); dp[i] = Math.max(dp[i], j * dp[i - j], j * (i - j));
} }
} }

24
problems/1005.K次取反后最大化的数组和.md
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@ -190,23 +190,23 @@ func largestSumAfterKNegations(nums []int, K int) int {
### Javascript ### Javascript
```Javascript ```Javascript
var largestSumAfterKNegations = function(nums, k) { var largestSumAfterKNegations = function(nums, k) {
nums.sort((a, b) => {
return Math.abs(b) - Math.abs(a) nums.sort((a,b) => Math.abs(b) - Math.abs(a))
})
for(let i = 0; i < nums.length; i++) { for(let i = 0 ;i < nums.length; i++){
if(nums[i] < 0 && k > 0) { if(nums[i] < 0 && k > 0){
nums[i] *= -1 nums[i] = - nums[i];
k-- k--;
} }
} }
if(k > 0 && k % 2 === 1) { // 若k还大于0,则寻找最小的数进行不断取反
nums[nums.length - 1] *= -1 while( k > 0 ){
nums[nums.length-1] = - nums[nums.length-1]
k--;
} }
k = 0
return nums.reduce((a, b) => { return nums.reduce((a, b) => {
return a + b a + b
}) })
}; };