Search Insert Position &Search in Rotated Sorted Array

class Solution {public:    int searchInsert(int A[], int n, int target) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if (!A) return 0;        if (target<A[0]) {            return 0;        }        if (target>A[n-1]) {            return n;        }        for (int i=0; i<n; i++) {            if (target==A[i]) {                return i;            }else if (target>A[i] && target < A[i+1]) {                return i+1;            }        }    }};

14/1/30: 二分法

class Solution {public:    int searchInsert(int A[], int n, int target) {        if (!n) return 0;        int i=0, j=n-1;        while(i<=j) {            int mid=(i+j)/2;            if (A[mid]==target) {                return mid;            }            if (A[mid]<target && A[mid+1]>target) {                return mid+1;            }            if (A[mid]<target) {                i=mid+1;            }else {                j=mid-1;            }        }        if (i==j) {            if (A[i]<target) {                return j+1;            }else {                return j;            }        }    }};

刚开始写的比较长，后来发现还可以化简：

class Solution {public:    int searchInsert(int A[], int n, int target) {        if (!n) return 0;        int i=0, j=n-1;        while(i<=j) {            int mid=(i+j)/2;            if (A[mid]==target) {                return mid;            }            if (A[mid]<target) {                i=mid+1;            }else {                j=mid-1;            }        }        return i;    }};

Search in Rotated Sorted Array

class Solution {public:    int search(int A[], int n, int target) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if (A[0]<target) {            for (int i=1; i<n; i++) {                if (A[i]==target) {                    return i;                }            }        }else if (A[0]>target) {            for (int i=n-1; i>0; i--) {                if (A[i]==target) {                    return i;                }            }        }else if (A[0]==target) {            return 0;        }        return -1;    } };

other better solution:

http://blog.unieagle.net/2012/10/23/leetcode%E9%A2%98%E7%9B%AE%EF%BC%9Asearch-in-rotated-sorted-array/

class Solution {public:    int search(int A[], int n, int target) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if(n <= 0) return -1;        int i = 0, j = n-1;        while(i <= j) {            int mid = (i + j) / 2;            int mval = A[mid];            if(target == mval) return mid;            if(i == j) return -1;            if(A[i] < A[j]) {                //we are in the normal sorted array range                if(target < mval)                    j = mid - 1;                else                    i = mid + 1;            } else if (A[i] > A[j]) {                //we are in the rotated range of array                if(mval > A[i]) {                    //mid is in the ascending part of arr                    if(target < mval) {                        if(target < A[i]) {                            //target should be in the rear of array                            i = mid + 1;                        } else if (target == A[i]) {                            return i;                        } else {                            //target should be in the front                            j = mid - 1;                        }                    }                    else                        i = mid + 1;                } else if (mval < A[i]) {                    if(target < mval) {                        j = mid - 1;                    } else {                        if(target == A[j])                            return j;                        else if(target < A[j])                            i = mid + 1;                        else                            j = mid - 1;                    }                } else {                    i = mid + 1;                }            } else {                //unsure                return -3;            }        }        return -1;    }};