1.1 KiB
1.1 KiB
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];
}
};