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查找算法.md

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    public static int binSearch(int[] arr, int key) {
        int left = 0;
        int right = arr.length - 1;
        int mid = 0;
        while (left <= right) {
            // 如果left与right都超过了int最大值的1/2,那么(left+right)会发生溢出,
            // 因此不能使用(left+right)/2求mid,而是应该用left+(right-left)/2
            mid = left + (right - left) / 2;
            if (key < arr[mid]) {
                right = mid - 1;
            } else if (key > arr[mid]) {
                left = mid + 1;
            } else {
                return mid;
            }
        }
        return -1;
    }
    /**
     * 解法1:遍历数组,查找缺失的数字
     */
    public int missingNumber(int[] nums) {
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] != i) {
                return i;
            }
        }
        return nums.length;
    }

    /**
     * 解法二:使用二分法查找丢失数字
     */
    public int missingNumberBinarySearch(int[] nums) {
        int left = 0, right = nums.length - 1;
        while (left <= right) {
            int mid = left + (left - right) / 2;
            if (nums[mid] == mid) { // 说明缺失的数字不在前半段数组
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        return left;
    }
    public int searchInsert(int[] nums, int target) {
        int left = 0, right = nums.length - 1, mid;
        while (left <= right) {
            mid = left + (right - left) / 2;
            if (target < nums[mid]) {
                right = mid - 1;
            } else if (target > nums[mid]) {
                left = mid + 1;
            } else {
                return mid;
            }
        }
        return left;
    }
    public int firstBadVersion(int n) {
        int left = 1, right = n, mid;
        while (left < right) {
            // 如果left与right都超过了int最大值的1/2,那么(left+right)会发生溢出,
            // 因此不能使用(left+right)/2求mid,而是应该用left+(right-left)/2
            mid = left + ((right - left) >> 1);
            if (isBadVersion(mid)) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }