题一.K次取反后最大化的数组和(LeetCode)
题目分析
由于我们希望数组的和尽可能大,因此除非万不得已,我们应当总是修改负数,并且优先修改值最小的负数。因为将负数 −x 修改成 x 会使得数组的和增加 2x,所以这样的贪心操作是最优的。
题解
class Solution {
public:
int largestSumAfterKNegations(vector<int>& nums, int k) {
int m = 0;
int ret = 0;
for (auto x : nums)
{
if (x < 0) m++;
}
sort(nums.begin(), nums.end());
if (k > m)
{
for (int i = 0; i < m; i++)
{
nums[i] *= -1;
}
sort(nums.begin(), nums.end());
for (int i = 0; i < k - m; i++)
{
nums[0] *= -1;
}
}
else
{
for (int i = 0; i < k; i++)
{
nums[i] *= -1;
}
}
for (auto x : nums)
{
ret += x;
}
return ret;
}
};
题二.按身高排序(LeetCode)
题解一.哈希表
class Solution1 {
public:
vector<string> sortPeople(vector<string>& names, vector<int>& heights) {
map<int, string> hash;
vector<string> ans;
for (int i = 0; i < names.size(); ++i) hash[heights[i]] = names[i];
for (auto &i : hash) ans.push_back(i.second);
reverse(ans.begin(), ans.end());
return ans;
}
};
*题解二.对下标排序
class Solution2 {
public:
vector<string> sortPeople(vector<string>& names, vector<int>& heights) {
int n = names.size();
vector<int> index(n);
for (int i = 0; i < n; i++) index[i] = i;
sort(index.begin(), index.end(), [&](int i, int j)
{
return heights[i] > heights[j];
});
vector<string> ret;
for (int i : index)
{
ret.push_back(names[i]);
}
return ret;
}
};
**题三.优势洗牌(LeetCode)
题目分析
题解
class Solution {
public:
vector<int> advantageCount(vector<int>& nums1, vector<int>& nums2) {
int n=nums1.size();
sort(nums1.begin(),nums1.end());
vector<int> index2(n);
for(int i=0;i<n;i++) index2[i]=i;
sort(index2.begin(),index2.end(),[&](int i,int j)
{
return nums2[i]<nums2[j];
});
vector<int> ret(n);
int left=0,right=n-1;
for(auto x : nums1)
{
if(x>nums2[index2[left]]) ret[index2[left++]]=x;
else ret[index2[right--]]=x;
}
return ret;
}
};