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<>题目 :二分查找
给定一个 n 个元素有序的(升序)整型数组 nums 和一个目标值 target ,写一个函数搜索 nums 中的
target,如果目标值存在返回下标,否则返回 -1。
示例 1:
输入: nums = [-1,0,3,5,9,12], target = 9
输出: 4
解释: 9 出现在 nums 中并且下标为 4
示例 2:
输入: nums = [-1,0,3,5,9,12], target = 2
输出: -1
解释: 2 不存在 nums 中因此返回 -1
<>二分查找的使用条件很简单
有序的排列即可
解题思路 :设定左右指针
找出中间位置,并判断该位置值是否等于 target
nums[mid] == target 则返回该位置下标
nums[mid] > target 则右侧指针移到中间
nums[mid] < target 则左侧指针移到中间
图解
<>有序数组用二分和不用二分的区别
直接遍历全部,效率只超过了百分之十
用了二分
差距十分的明显,算法之妙
解题代码
public static 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) { return mid; } else if(nums[mid] > target) { right = mid - 1; }
else { left = mid + 1; } } return -1; }