实现二分查找

用二分实现的关键点：

• 凡有序，必二分
• 凡二分，时间复杂度比包含O(logn)
• 可以用递归 和 非递归 两种方式实现
  • 递归实现的二分，逻辑更清晰一些
  • 用非递归实现，性能更好一些

代码实现：　

• 非递归实现二分法

  export function binarySearch1(arr:number[],target: number): number{
      const length = arr.length;
      if(length === 0) return -1
  
      let startIndex = 0;  //开始位置
      let endIndex = length -1;//结束位置
  
      while(startIndex <= endIndex){
          const midIndex = Math.floor((endIndex + startIndex) / 2)
          const midValue = arr[midIndex]
          if(target < midValue){
              endIndex = midIndex -1 
          }else if(target > midValue){
              startIndex = midIndex + 1
          }else {
              return midIndex
          }
      }
      return -1
  }

• 递归实现二分法

  export function binarySearch2 (arr: number[],target: number, startIndex?: number,endIndex?:number){
      const length = arr.length;
      if(length === 0){
          return -1
      }
      startIndex = startIndex || 0
      endIndex = endIndex || length -1
      // 如果 start 大于 end ,则结束
      if(startIndex > endIndex){
          return -1
      }
  
      const midIndex = Math.floor((startIndex + endIndex) / 2)
      const midValue = arr[midIndex]
      if(target < midValue){ //目标较小，则继续查找
          return binarySearch2(arr,target,startIndex,midIndex -1)
      }else if(target > midValue){ //目标较大，则继续查找
          return binarySearch2(arr,target,midIndex + 1,endIndex)
      }else{
          // 相等，返回
          return midIndex
      }
  }

• 测试两种实现的测试用例

  /**
   * @description 二分查找测试
   */
  import  {binarySearch1,binarySearch2} from './binary-search'
  
  describe('二分查找测试',() =>{ 
      it('binarySearch1正常查找', ()=>{
          const arr = [20,30,40,50,60,70]
          const res = binarySearch1(arr,60)
          expect(res).toBe(4)
      })
      it('binarySearch1空数组', ()=>{
          const arr: number[] = []
          const res = binarySearch1(arr,60)
          expect(res).toBe(-1)
      })
      it('binarySearch1不存在情况', ()=>{
          const arr = [20,30,40,50,60,70]
          const res = binarySearch1(arr,45)
          expect(res).toBe(-1)
      })
  
      it('binarySearch2正常查找', ()=>{
          const arr = [20,30,40,50,60,70]
          const res = binarySearch2(arr,60)
          expect(res).toBe(4)
      })
      it('binarySearch2空数组', ()=>{
          const arr: number[] = []
          const res = binarySearch2(arr,60)
          expect(res).toBe(-1)
      })
      it('binarySearch2不存在情况', ()=>{
          const arr = [20,30,40,50,60,70]
          const res = binarySearch2(arr,45)
          expect(res).toBe(-1)
      })
  })