题解 | #二分查找-II#

/**
     * 当查找到满足条件的值且是从左到右第一个的时候，right=mid-1,继续二分查找，
     * 会一直从left查找到nums[mid]=target的mid,
     * 1-----2 --->   1-----2      --->   1-----2   ---> 不满足条件返回a
     * a(mid)     b        ab(mid)        b     a 
     * 
     * @param nums
     * @param target
     * @return
     */
    public static int search1(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return -1;
        }
        int a = 0;
        int b = nums.length - 1;
        while (a <= b) {
            int mid = (a + b) / 2;
            if (nums[mid] > target) {
                b = mid - 1;
            } else if (nums[mid] < target) {
                a = mid + 1;
            } else {
                b = mid - 1;
            }
        }
        if (a >= nums.length || nums[a] != target) {
            return -1;
        }
        return a;
    }