lintcode 6. Merge Two Sorted Arrays（多回顾二分法）

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解法：很普通的三步走法，要主要加极端条件的判断；时间会大大缩短！！！
不加判断，只能打败4.4%；
加了之后 91%

class Solution {
public:
    /**
     * @param A: sorted integer array A
     * @param B: sorted integer array B
     * @return: A new sorted integer array
     */
    vector mergeSortedArray(vector &A, vector &B) {
        // write your code here
        vector result;
        if(A.empty()) return B;
        if(B.empty()) return A;
        int m = A.size(), i = 0;
        int n = B.size(), j = 0;
        while(i < m && j < n){
            if(A[i] <= B[j]){
                result.push_back(A[i++]);
            }
            else{
                result.push_back(B[j++]);
            }
        }
        if(i < m){
            while(i < m){
                result.push_back(A[i++]);
            }
        }
        else{
            while(j < n){
                result.push_back(B[j++]);
            }
        }
        return result;
    }
};

！！！解法er: 优化，当A很长，B很短的时候，可以用二分法，将B中的元素插入到A中。

class Solution {
public:
    /**
     * @param A: sorted integer array A
     * @param B: sorted integer array B
     * @return: A new sorted integer array
     */
    vector mergeSortedArray(vector &A, vector &B) {
        // write your code here
        vector result;
        int m = A.size();
        int n = B.size();
        int pa = 0;
        for(int i = 0; i < n; i++){
            int index = binarySearch(A,B[i]);
            cout< nums, int target) {
        int left = 0;
        int right = nums.size() - 1;
        while(left <= right){
            int middle = (left + right)  / 2;
            if(nums[middle] == target) return middle;
            if(nums[middle] > target){
                right = middle - 1;
            }
            else{
                left = middle + 1;
            }
        }
        return left;
    }
};