First Missing Positive
Given an unsorted integer array, find the first missing positive integer.
For example,
Given [1,2,0]
return 3
,
and [3,4,-1,1]
return 2
.
Your algorithm should run in O(n) time and uses constant space.
关键是O(n)复杂度及常量运行空间。
采用交换法。
class Solution { public: int firstMissingPositive(int A[], int n) { for(int i=0;i<n;) { if(A[i]>0&&A[i]<=n&&A[i]!=i+1&&A[A[i]-1]!=A[i]) { swap(A[A[i]-1],A[i]); } else i++; } for(int i=0;i<n;i++) { if(A[i]!=i+1) { return i+1; } } return n+1; } };
采用另外一种hash方法。某大牛想出来的:
- 第一遍扫描排除所有非正的数,将它们设为一个无关紧要的正数(n+2),因为n+2不可能是答案
- 第二遍扫描,将数组作为hash表来使用,用数的正负来表示一个数是否存在在A[]中。
当遇到A[i],而A[i]属于区间[1,n],就把A中位于此位置A[i] – 1的数置翻转为负数。 所以我们取一个A[i]的时候,要取它的abs,因为如果它是负数的话,通过步骤一之后,只可能是我们主动设置成负数的。 - 第三遍扫描,如果遇到一个A[i]是正数,说明i+1这个数没有出现在A[]中,只需要返回即可。
- 上一步没返回,说明1到n都在,那就返回n+1
class Solution { public: int firstMissingPositive(int A[], int n) { if(n <= 0) return 1; const int IMPOSSIBLE = n + 1; for(int i = 0; i < n; ++i) { if(A[i] <= 0) A[i] = IMPOSSIBLE; } for(int i=0; i < n; ++i) { int val = abs(A[i]); if(val <= n && A[val-1] > 0) A[val-1] *= -1; } for(int i = 0; i<n; ++i) { if(A[i] > 0) return i+1; } return n+1; } };
覆盖区间
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: vector<Interval> insert(vector<Interval> &intervals, Interval newInterval) { vector<Interval> result; vector<Interval>::iterator it; bool flag=true; for (it=intervals.begin();it!=intervals.end();it++) { if (it->end<newInterval.start) { result.push_back(*it); continue; } if (it->start>newInterval.end) { if (flag) { result.push_back(newInterval); flag=false; } result.push_back(*it); continue; } newInterval.start=it->start<newInterval.start?it->start:newInterval.start; newInterval.end=it->end>newInterval.end?it->end:newInterval.end; } if (flag) { result.push_back(newInterval); } return result; } };