47 lines
1.1 KiB
Markdown
47 lines
1.1 KiB
Markdown
|
||
和之前个回溯法的各个总和串起来一波，本题用回溯稳稳的超时
|
||
|
||
```
|
||
class Solution {
|
||
public:
|
||
int combinationSum4(vector<int>& nums, int target) {
|
||
vector<int> dp(target + 1, 0);
|
||
dp[0] = 1;
|
||
for (int i = 0; i <= target; i++) {
|
||
for (int j = 0; j < nums.size(); j++) {
|
||
if (i - nums[j] >= 0) {
|
||
dp[i] += dp[i - nums[j]];
|
||
}
|
||
}
|
||
}
|
||
return dp[target];
|
||
}
|
||
};
|
||
```
|
||
|
||
|
||
C++测试用例有超过两个树相加超过int的数据
|
||
超限的情况
|
||
|
||
|
||
一些题解会直接用ull usigned long long
|
||
|
||
java 也是四个字节，理论上没有差别，可能后台java和C++测试用例不同，bug.....
|
||
```
|
||
class Solution {
|
||
public:
|
||
int combinationSum4(vector<int>& nums, int target) {
|
||
vector<int> dp(target + 1, 0);
|
||
dp[0] = 1;
|
||
for (int i = 0; i <= target; i++) {
|
||
for (int num : nums) {
|
||
if (i - num >= 0 && dp[i] < INT_MAX - dp[i - num]) {
|
||
dp[i] += dp[i - num];
|
||
}
|
||
}
|
||
}
|
||
return dp[target];
|
||
}
|
||
};
|
||
```
|