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1. praemittit et artes

(1) Praefixum sum

Praeposita ars summa celeriter et saepe apta est ad summam elementorum intra indicem range computandum.

class NumArray {
public:
    vector<int> preSum; //前缀和数组
    NumArray(vector<int>& nums) {
        //preSum[0] = 0，便于计算累加和
        preSum.resize(nums.size() + 1, 0);
        for(int i = 0; i < nums.size(); i++){
            preSum[i + 1] = preSum[i] + nums[i];
        }
    }
    
    int sumRange(int left, int right) {
        // 查询闭区间 [left, right] 的累加和
        return preSum[right + 1] - preSum[left];
    }
};

(2) Prefix sum + Nullam mensam

Praefixio summa ordinata adiuvat ut cito summam elementorum subarrayorum computaveris, sed quid si rogaris ut subarrayas condiciones obvias invenias?For example, let younumsVultus concordiaetarget subarray, etiam ope praefixionum et vestium, adhuc pro ansa nidificata scribere debes.Sed uti possumus tabulam anticam ad singulas praepositionis indices et correspondentes notare, ut celeriter summam scopo computare possimus.targetde subarray

class Solution {
public:
    int findMaxLength(vector<int>& nums) {
        int len = 0;
        vector<int> preSum(nums.size() + 1, 0);
 
        for(int i = 0; i < nums.size(); i++)
            preSum[i+1] = preSum[i] + (nums[i] == 0 ? -1 : 1);
 
        // 前缀和到索引的映射，方便快速查找所需的前缀和
        unordered_map<int, int> umap;
 
        for(int i = 0; i < preSum.size(); i++){
            // 如果这个前缀和还没有对应的索引，说明这个前缀和第一次出现，记录下来
            if(umap.find(preSum[i]) == umap.end()) umap[preSum[i]] = i;
            else len = max(len, i - umap[preSum[i]]);
        }
        return len;
    }
};

class Solution {
public:
    bool checkSubarraySum(vector<int>& nums, int k) {
        vector<int>preSum(nums.size() + 1, 0);
 
        for(int i = 0; i < nums.size(); i++)
            preSum[i + 1] = preSum[i] + nums[i];
 
        unordered_map<int, int> umap;
 
        // 寻找 i, j 使得 (preSum[i] - preSum[j]) % k == 0 且 i - j >= 2
        // (preSum[i] - preSum[j]) % k == 0 其实就是 preSum[i] % k == preSum[j] % k
        for(int i = 0; i < preSum.size(); i++){
            auto it = umap.find(preSum[i] % k);
            if(it == umap.end()) umap[preSum[i] % k] = i;
            else if ((i - it->second) >=2) return true;
        }
        return false;
    }
};

class Solution {
public:
    int subarraySum(vector<int>& nums, int k) {
        int ans = 0;
        vector<int> preSum(nums.size() + 1, 0);
 
        for(int i = 0; i < nums.size(); i++){
            preSum[i + 1] = preSum[i] + nums[i];
        }
 
        // 寻找 i, j 使得 preSum[i] - preSum[j] == k, i > j
        // nums = [1,2,3], k = 3, preSum = [0,1,3,6]
        unordered_map<int, int> umap;
        for(int i = 0; i < preSum.size(); i++){
            if(umap.find(preSum[i] - k) != umap.end()) ans += umap[preSum[i] - k]; // 该语句必须放在前面
            
            if(umap.find(preSum[i]) == umap.end()) umap[preSum[i]] = 1;
            else umap[preSum[i]]++;
        }
        return ans;
 
    }
};