二分查找模板

二分查找是建立在有序的基础上的

基础模板

int bsearch(vector<int>& numbers, int value) {
    int left=0, right = numbers.size()-1;
    while(left <= right){ //有的写法这里是 < , 则条件结束时返回的需是 left 而非 -1
        int mid = left + (right - left)/2;//防止溢出
        if(numbers[mid] == value){
            return mid;
        }else if(numbers[mid] < value){
            left = mid + 1;
        }else{
            right = mid - 1;
        }
    }
    return -1;
}

变种

1.查找第一个值等于给定值的元素

int bsearch(vector<int>& numbers, int value) {
    int left=0, right = numbers.size()-1;
    while(left <= right){ //有的写法这里是 < , 则条件结束时返回的需是 left 而非 -1
        int mid = left + (right - left)/2;//防止溢出
        if(numbers[mid] < value){
            left = mid + 1;
        }else if(numbers[mid] > value){
            right = mid - 1;
        }else{
            //1.值相等 2.前一个值不是value
            if(mid==0 || numbers[mid-1]!=value){
                  return mid;
            }else{
                  right = mid - 1;
            }
        }
    }
    return -1;
}

2.查找最后一个值等于给定值的元素

int bsearch(vector<int>& numbers, int value) {
    int left=0, right = numbers.size()-1;
    while(left <= right){ //有的写法这里是 < , 则条件结束时返回的需是 left 而非 -1
        int mid = left + (right - left)/2;//防止溢出
        if(numbers[mid] < value){
            left = mid + 1;
        }else if(numbers[mid] > value){
            right = mid - 1;
        }else{
            //1.值相等 2.后一个值不是value
            if(mid==numbers.size()-1 || numbers[mid+1]!=value){
                  return mid;
            }else{
                  left = mid + 1;
            }
        }
    }
    return -1;
}

3.查找第一个值大于等于给定值的元素

int bsearch(vector<int>& numbers, int value) {
    int left=0, right = numbers.size()-1;
    while(left <= right){ //有的写法这里是 < , 则条件结束时返回的需是 left 而非 -1
        int mid = left + (right - left)/2;//防止溢出
        if(numbers[mid] < value){
            left = mid + 1;
        }else{
            //1.>= 2.前一个值小于value
            if(mid==0 || numbers[mid-1]<value){
                  return mid;
            }else{
                  right = mid - 1;
            }
        }
    }
    return -1;
}

4.查找最后一个值小于等于给定值的元素

int bsearch(vector<int>& numbers, int value) {
    int left=0, right = numbers.size()-1;
    while(left <= right){ //有的写法这里是 < , 则条件结束时返回的需是 left 而非 -1
        int mid = left + (right - left)/2;//防止溢出
        if(numbers[mid] > value){
            right = mid - 1;
        }else{
            //1.<= 2.后一个值大于value
            if(mid==numbers.size()-1 || numbers[mid+1]>value){
                  return mid;
            }else{
                  left = mid + 1;
            }
        }
    }
    return -1;
}