題目：

Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.

You may assume NO duplicates in the array.

[1,3,5,6]，5 → 2

[1,3,5,6]，2 → 1

[1,3,5,6]， 7 → 4

[1,3,5,6]，0 → 0

題解：

Solution 1 ()

class Solution {
    
 
/** 
     * param A : an integer sorted array
     * param target :  an integer to be inserted
     * return : an integer
     */
public:
    int searchInsert(vector<int> &A, int target) {
        if (A.size() == 0) {
            return 0;
        }
        int start = 0;
        int end = A.size() - 1
 
;
        
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (A[mid] == target) {
                return mid;
            } else if (A[mid] > target) {
                end = mid;
            } else {
                start = mid;
            }
        }
        
 
if (A[start] >= target) {
            return start;
        } else if (A[end] >= target) {
            return end;
        } else {
            return end + 1;
        }
    }
};

Solution 2 ()

class Solution {
    /** 
     * param A : an integer sorted array
     * param target :  an integer to be inserted
     * return : an integer
     */
public:
    int searchInsert(vector<int> &A, int target) {
        if (A.size() == 0) {
            return 0;
        }
        int start = 0;
        int end = A.size() - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (A[mid] == target) {
                return mid;
            } else if (A[mid] > target) {
                end = mid;
            } else {
                start = mid;
            }
        }
        if (A[end] == target) {
            return end;
        } 
        if (A[end] < target) {
            return end + 1;
        } 
        if (A[start] >= target) {
            return start;
        }
        
        return start + 1;
    }
};

【LintCode】060.Search Insert Position