题解 | #二分查找-I#

题解里的二分查找都是比较mid和target之后，left+1或right-1，这样查找的效率感觉不高啊，如果直接将left = mid或right = mid，就直接切一半来找，感觉要快一点。

/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param nums int整型一维数组 
 * @param target int整型 
 * @return int整型
 */
function search( nums ,  target ) {
    // write code here
    let left = 0, right = nums.length-1;
    let mid = Math.floor((left + right)/2);
    if (nums.length == 0 || (nums.length == 1 && nums[0] != target)) {
        return -1;
    } else if (nums.length == 1 && nums[0] == target) {
        return 0;
    }
    while (right - left >= 1) {
        if (right - left == 1 && nums[right] == target) {
            return right;
        } 
        if (target == nums[mid]) {
            return mid
        } else if (target > nums[mid]) {
            if (right - left == 1 && nums[right] != target) {
            return -1;
        }
            left = mid
        } else {
            right = mid;
        }
       mid = Math.floor((left + right)/2);
    }
    return -1
}
module.exports = {
    search : search
};