220. Contains Duplicate III—medium

Given an array of integers, find out whether there are two distinct indices i and j in the array such that the difference between nums[i] andnums[j] is at most t and the difference between i and j is at most k.
题目解释：并不是要求所有，只要有一组满足即可
暴力法：（超时）

class Solution {public:    bool containsNearbyAlmostDuplicate(vector<int>& nums, int k, int t) {        if (nums.size() <= 1 || k == 0)            return false;        vector<long int> numDiff;        vector<long int> sumVec(nums.size());        sumVec[0] = 0;        for (int i = 0; i < nums.size() - 1; i ++){            numDiff.push_back(nums[i+1] - nums[i]);     //nums[j] - nums[i] = numDiff[i] + numDiff[i+1] + ... + numDiff[j-1]        }        for(int i = 0; i < nums.size(); i ++){            for (int len = 1; len <= k && i + len < nums.size(); len ++){                sumVec[len] = sumVec[len - 1] + numDiff[i+len-1];                if(abs(sumVec[len]) <= t)                    return true;            }        }        return false;    }};

下面的借鉴的一种，最差复杂度也是O(N^2)，但是出现最差情况的几率很小

class Solution {public:    static bool lessthan(pair<int, int> a, pair<int, int> b){        return a.first < b.first;    }    bool containsNearbyAlmostDuplicate(vector<int>& nums, int k, int t) {        vector<pair<int, int> > a;        for (int i = 0; i < nums.size(); i ++){            a.push_back(make_pair(nums[i], i));        }        sort(a.begin(), a.end(), lessthan);        int big = 0, small = 0;        for(int big = 1; big < nums.size(); big ++){            while((long)a[big].first - a[small].first > t) small ++;            for(int j = small; j < big; j ++){                if (abs(a[big].second - a[j].second) <= k){                    return true;                }            }        }        return false;    }};