先贴代码：

public int searchInsert(int[] nums, int target) {
        int i = 0;
        int j = nums.length - 1;
        while (i <= j) {
            int mid = i + (j - i) / 2;
            if (nums[mid] < target) {
                i = mid + 1;
            } else if (nums[mid] > target) {
                j = mid - 1;
            } else if (nums[mid] == target) {
                return mid;
            }
        }
        return i;
    }

一定要注意！ int mid = i + (j - i) / 2 不能写成 int mid = (i + j) / 2 , 不然会越界。

放一道例题

补充：当数组中有重复数，且需要满足一个条件，返回满足的子串长度等情况，下面这个二分也适用

class Solution {
    public int[] successfulPairs(int[] spells, int[] potions, long success) {
        Arrays.sort(potions);
        int[] res = new int[spells.length];

        for (int i = 0; i < spells.length; i++) {
            int left = 0, right = potions.length - 1;
            int index = potions.length;
            while (left <= right) {
                int mid = (right - left) / 2 + left;
                if ((long)spells[i] * potions[mid] >= success) {
                    right = mid - 1;
                    index = mid;
                } else {
                    left = mid + 1;
                }
            }
            res[i] = potions.length - index;
        }
        return res;
    }
}