力扣hot100——普通数组

53. 最大子数组和

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int n = nums.size();
        int ans = nums[0];
        int sum = 0;
        for (int i = 0; i < n; i++) {
            sum += nums[i];
            ans = max(ans, sum);
            if (sum < 0) sum = 0;
        }
        return ans;
    }
};

最大子数组和，贪心即可

 56. 合并区间

class Solution {
public:
    vector<vector<int>> merge(vector<vector<int>>& intervals) {
        vector<vector<int>> ans;
        sort(intervals.begin(), intervals.end());
        int L = -1, R = -1;
        for (auto vec: intervals) {
            int l = vec[0], r = vec[1];
            if (l > R) {
                if (R != -1) {
                    ans.push_back({L, R});
                }
                L = l, R = r;
            } 
            R = max(R, r);
        }
        ans.push_back({L, R});
        return ans;
    }
};

排序，贪心。区间合并板子题

189. 轮转数组

class Solution {
public:
    void rotate(vector<int>& nums, int k) {
        vector<int> v;
        int n = nums.size();
        k %= n;
        for (int i = n - 1; i >= n - k; i--) {
            v.push_back(nums[i]);
        }
        reverse(v.begin(), v.end());
        for (int i = 0; i <= n - k - 1; i++) {
            v.push_back(nums[i]);
        }
        swap(nums, v);
    }
};

模拟，注意k可能大于数组大小即可 

 238. 除自身以外数组的乘积

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        int n = nums.size();
        vector<int> a(n + 10, 0);
        for (int i = 1; i <= n; i++) a[i] = nums[i - 1];
        vector<int> pre(n + 10, 0);
        vector<int> suf(n + 10, 0);
        pre[0] = 1;
        suf[n + 1] = 1;
        for (int i = 1; i <= n; i++) {
            pre[i] = pre[i - 1] * a[i];
        }

        for (int i = n; i >= 1; i--) {
            suf[i] = suf[i + 1] * a[i];
        }
        
        vector<int> ans;
        for (int i = 1; i <= n; i++) {
            int x = pre[i - 1] * suf[i + 1];
            ans.push_back(x);
        }
        return ans;
    }
};

前缀积和后缀积

41. 缺失的第一个正数

class Solution {
public:
    int firstMissingPositive(vector<int>& nums) {
        int n = nums.size();
        int max = n + 1;

        for (int i = 0; i < n; i++) {
            if (nums[i] <= 0 || nums[i] >= max) {
                nums[i] = max;
            }
        }

        for (int i = 0; i < n; i++) {
            int t = abs(nums[i]);
            if (t <= n) {
                nums[t - 1] = -abs(nums[t - 1]);
            }
        }

        for (int i = 0; i < n; i++) {
            if (nums[i] > 0) {
                return i + 1;
            }
        }
        return max;
    }
};

贪心结论，答案范围是[1, n + 1]
再通过一些技巧标记原数组表示哪个数存在