草稿图网站
java的Deque

Leetcode 704. 二分查找

题目：704. 二分查找

解题思路

1.左闭右闭区间的搜索，循环条件为left <= right。

2.左闭右开区间的搜索，循环条件为left < right。

代码

//1.左闭右闭
class Solution {
	public int search(int[] nums, int target) {
		int left = 0, right = nums.length - 1;
		while (left <= right){
			int mid = left + (right - left) / 2;
			if (nums[mid] > target){
				right = mid - 1;
			}else if (nums[mid] < target){
				left = mid + 1;
			}else{
				return mid;
			}
		}
		return -1;

	}
}
//2.左闭右开
class Solution {
	public int search(int[] nums, int target) {
		int left = 0, right = nums.length;
		while (left < right){
			int mid = left + (right - left) / 2;
			if (nums[mid] > target){
				right = mid;
			}else if (nums[mid] < target){
				left = mid + 1;
			}else{
				return mid;
			}
		}
		return -1;

	}
}

总结

暂无

Leetcode 27. 移除元素

题目：27. 移除元素

解题思路

双指针，快指针找到第一个不是val的后，两个指针一起走一步

代码

class Solution {
    public int removeElement(int[] nums, int val) {
        int left = 0, right = 0;
        while (right < nums.length){
            while (right < nums.length && nums[right] == val){
                right++;
            }
            if (right >= nums.length)
                break;
            nums[left++] = nums[right++];
        }
        return left;
    }
}

总结

代码有一定冗余，karl哥的代码更精炼

class Solution {
    public int removeElement(int[] nums, int val) {
        int slowIdx = 0;
        for (int fastIdx = 0; fastIdx < nums.length; fastIdx++){
            if (val != nums[fastIdx]){
                nums[slowIdx++] = nums[fastIdx];
            }
        }
        return slowIdx;
    }
}