Maximum Subarray

Find the contiguous subarray within an array (containing at least one number) which has the largest sum.

For example, given the array [−2,1,−3,4,−1,2,1,−5,4],
the contiguous subarray [4,−1,2,1] has the largest sum = 6.

class Solution {
public:
    int maxSubArray(int A[], int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int maxSum = INT_MIN;
        int sum = 0;
        for(int i = 0; i < n; i++){
            if(sum + A[i] < 0){
                sum = 0;
                maxSum = max(maxSum, A[i]);            
            }else{
                sum = sum + A[i];
                maxSum = max(maxSum, sum);
            }
            
        }
        return maxSum;
    }
};

Insert Interval

Given a set of non-overlapping intervals, insert a new interval into the intervals (merge if necessary).

You may assume that the intervals were initially sorted according to their start times.

Example 1:
Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].

Example 2:
Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16].

This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */

class Solution {
public:
    bool hasOverlap(Interval &a, Interval &b){
        if(a.start > b.end || b.start > a.end){
            return false;
        }
        return true;
    }

    vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<Interval> ret;
        ret.clear();        
        int size = intervals.size();
        bool flag_addNew = false;    
        for(int i = 0; i < size; i++){
            if(!flag_addNew && hasOverlap(intervals[i], newInterval)){
                newInterval.start = min(newInterval.start, intervals[i].start);
                newInterval.end = max(newInterval.end, intervals[i].end);
            }else{
                if(flag_addNew == false && (newInterval.start < intervals[i].start)){
                    ret.push_back(newInterval);
                    flag_addNew = true;
                }
                ret.push_back(intervals[i]);
            }
        }
        if(flag_addNew == false){
            ret.push_back(newInterval);
        }
        return ret;     
    }
};

Merge Intervals

Given a collection of intervals, merge all overlapping intervals.

For example,
Given [1,3],[2,6],[8,10],[15,18],
return [1,6],[8,10],[15,18].

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */



bool compare(const Interval &a, const Interval &b){
    return a.start < b.start;
}


class Solution {
public:

    vector<Interval> merge(vector<Interval> &intervals) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<Interval> ret;
        ret.clear();
        int size = intervals.size();
        if(size == 0)   return ret;
        sort(intervals.begin(), intervals.end(), mycompare);        
        Interval pre = intervals[0];
        for(int i = 1; i <= size; i++){
            if(i == size){
                ret.push_back(pre);
                break;
            }
            Interval cur = intervals[i];
            if(pre.end >= cur.start){
                pre.end = max(cur.end, pre.end);//error: forget max
            }else{
                ret.push_back(pre);
                pre = cur;
            }
        }
        return ret;        
    }
};